Question 11 Mark
A student finds 7 books of his interest but can borrow only three books. He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed. Find the number of ways he can choose three books that he wants to borrow.
Question 21 Mark
Ten students are to be selected for a project from a class of 30 students. There are 4 students who want to be together either in the project or not in the project. Find the number of possible selections.
Question 31 Mark
English alphabet has 11 symmetric letters that appear the same when looked at in a mirror. These letters are $A, H, I, M, O, T, U, V, W, X$, and $Y$. How many symmetric three letters passwords can be formed using these letters?
AnswerNumber of 3 Letter passwords $={ }^{11} \mathrm{P}_3$
$
\begin{aligned}
& =11 \times 10 \times 9 \\
& =990
\end{aligned}
$
View full question & answer→Question 41 Mark
Four parallel lines intersect another set of five parallel lines. Find the number of distinct parallelograms that can be formed.
AnswerWe need 2 lines from each set.
Required number $={ }^4 \mathrm{C}_2 \times{ }^5 \mathrm{C}_2$
$
\begin{aligned}
& =6 \times 10 \\
& =60
\end{aligned}
$
View full question & answer→Question 51 Mark
How many words can be formed by writing letters in the word CROWN in a different order?
AnswerFive Letters of the word CROWN are to be permuted.
Number of different words $=5 !=120$
View full question & answer→Question 61 Mark
There are 8 doctors and 4 lawyers in a panel. Find the number of ways for selecting a team of 6 if at least one doctor must be in the team.
AnswerThere are 8 doctors and 4 lawyers.
We need to select a team of 6 which contains at least one doctor.
Since, there are only 4 lawyers any team of 6 will contain at least two doctors.
Required number $={ }^{12} \mathrm{C}_6=924$
View full question & answer→Question 71 Mark
Find the number of ways of dividing 20 objects in three groups of sizes 8,7 and 5.
AnswerSelect 8 object out of 20 in ${ }^{20} C_8$ ways
Select 7 object from remaining 12 in ${ }^{12} C_7$ ways and 5 objects from remaining 5 in ${ }^5 C_5$ ways
Required number is $={ }^{20} \mathrm{C}_8 \times{ }^{12} \mathrm{C}_7 \times{ }^5 \mathrm{C}_5$
View full question & answer→Question 81 Mark
A question paper has 6 questions. How many ways does a student have if he wants to solve at least one question?
AnswerEvery question is 'SOLVED' or 'NOT SOLVED'.
There are 6 question.
Number of outcomes $=2^6$
This number includes the case when the student solves NONE of the question.
Required number $=2^6-1$
$
\begin{aligned}
& =64-1 \\
& =63
\end{aligned}
$
View full question & answer→Question 91 Mark
How many six-digit telephone numbers can be formed if the first two digits are 45 and no digit can appear more than once?
Question 101 Mark
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficients if a coefficient can be repeated in an equation?
Question 111 Mark
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
AnswerEvery lamp is either ON or OFF.
There are 12 lamps
Number of instances $=2^{12}$
This number includes the case in which all 12 lamps are OFF.
$\therefore$ Required Number $=2^{12}-1=4095$
View full question & answer→Question 121 Mark
Nine friends decide to go for a picnic in two groups. One group decides to go by car and the other group decides to go by train. Find the number of different ways of doing so if there must be at least 3 friends in each group.
Question 131 Mark
A student passes an examination if he secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.
AnswerEvery subject a student may pass or fail.
$\therefore$ Total number of outcomes $=2^7=128$
This number includes one case when the student passes in all subjects.
$\therefore$ Required number $=128-1=127$
View full question & answer→Question 141 Mark
30 objects are to be divided into three groups containing 7, 10, 13 objects. Find the number of distinct ways of doing so.
AnswerRequired number $={ }^{30} \mathrm{C}_7 \times{ }^{23} \mathrm{C}_{10} \times{ }^{13} \mathrm{C}_{13}$
View full question & answer→Question 151 Mark
Find $x$ if ${ }^n P_r=x^n C_r$
Answer$
\begin{aligned}
& { }^n \mathrm{P}_{\mathrm{r}}=x\left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\right) \\
& x=\frac{{ }^n \mathrm{P}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}} \\
& =\frac{\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}}{\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r} !) \mathrm{r} !}} \\
& =\mathrm{r} !
\end{aligned}
$
View full question & answer→Question 161 Mark
Find $n$ if ${ }^n C_{n-2}=15$
Answer$
\begin{aligned}
& { }^n C_{n-2}=15 \\
& \therefore{ }^n C_2=15 \ldots \ldots\left[\because{ }^n C_r={ }^n C_{n-r}\right] \\
& \therefore \frac{n !}{(n-2) ! 2 !}=15 \\
& \therefore \frac{n(n-1)(n-2) !}{(n-2) ! 2 \times 1}=15 \\
& \therefore n(n-1)=30 \\
& \therefore n(n-1)=6 \times 5
\end{aligned}
$
Comparing both sides, we get
$
\therefore \mathrm{n}=6
$
View full question & answer→Question 171 Mark
Find $n$ if ${ }^{2 n} C_{r-1}={ }^{2 n} C_{r+1}$
Answer$
\begin{aligned}
& { }^{2 n} C_{r-1}={ }^{2 n} C_{r+1} \\
& \text { If }{ }^n C_x={ }^n C_y \text {, then either } x=y \text { or } x=n-y \\
& \therefore r-1=r+1 \text { or } r-1=2 n-(r+1)
\end{aligned}
$
But $r-1=r+1$ is not possible
$
\begin{aligned}
& \therefore \mathrm{r}-1=2 \mathrm{n}-(\mathrm{r}+1) \\
& \therefore \mathrm{r}+\mathrm{r}=2 \mathrm{n} \\
& \therefore \mathrm{r}=\mathrm{n}
\end{aligned}
$
View full question & answer→Question 181 Mark
Find $n$ if ${ }^{23} C_{3 n}={ }^{23} C_{2 n+3}$
Answer$
\begin{aligned}
& { }^{23} C_{3 n}={ }^{23} C_{2 n+3} \\
& \text { If }{ }^n C_x={ }^n C_y \text {, then either } x=y \text { or } x=n-y \\
& \therefore 3 n=2 n+3 \text { or } 3 n=23-2 n-3 \\
& \therefore n=3 \text { or } n=4
\end{aligned}
$
View full question & answer→Question 191 Mark
Find $n$ if ${ }^n C_8={ }^n C_{12}$
Answer$
{ }^n \mathrm{C}_8={ }^n \mathrm{C}_{12}
$
If ${ }^n C_x={ }^n C_y$, then either $x=y$ or $x=n-y$
$
\therefore 8=12 \text { or } 8=\mathrm{n}-12
$
But $8=12$ is not possible
$
\begin{aligned}
& \therefore 8=\mathrm{n}-12 \\
& \therefore \mathrm{n}=20
\end{aligned}
$
View full question & answer→Question 201 Mark
If ${ }^n P_r=1814400$ and ${ }^n C_r=45$, find $r$.
Answer$
\begin{array}{ll}
& { }^n \mathrm{P}_{\mathrm{r}}=1814400,{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=45 \\
\therefore \quad & \frac{{ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{1814400}{45} \\
\therefore \quad & \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !} \times \frac{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}{\mathrm{n} !}=\frac{1814400}{45} \\
\therefore \mathrm{r} != & 40320 \\
\therefore \mathrm{r} != & 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\
\therefore \mathrm{r} != & 8 ! \\
\therefore \mathrm{r}= & 8
\end{array}
$
View full question & answer→Question 211 Mark
Find n if: ${ }^n \mathrm{C}_{n-3}=84$
Answer$
\begin{aligned}
& { }^n C_{n-3}=84 \\
& \therefore \frac{n !}{(n-3) ![n-(n-3)] !}=84 \\
& \therefore \frac{n(n-1)(n-2)(n-3) !}{(n-3) ! \times 3 !}=84 \\
& \therefore n(n-1)(n-2)=84 \times 6 \\
& \therefore \mathrm{n}(n-1)(n-2)=9 \times 8 \times 7
\end{aligned}
$
Comparing on both sides, we get
$
\therefore \mathrm{n}=9
$
View full question & answer→Question 221 Mark
Find n if: ${ }^6 \mathrm{P}_2=\mathrm{n}{ }^6 \mathrm{C}_2$
Answer$
\begin{array}{ll}
& { }^6 \mathrm{P}_2=\mathrm{n}\left({ }^6 \mathrm{C}_2\right) \\
\therefore \quad & \frac{6 !}{(6-2) !}=\mathrm{n} \frac{6 !}{2 !(6-2) !} \\
\therefore \quad & \frac{6 !}{4 !}=\mathrm{n} \frac{6 !}{2 ! 4 !} \\
\therefore \quad & \mathrm{n}=2 !=2 \times 1=2
\end{array}
$
View full question & answer→Question 231 Mark
Find the number of triangles formed by joining 12 points if: four points are collinear
AnswerThere are 12 points on the plane: When 4 of these points are collinear:
$
\begin{aligned}
& \therefore \text { Number of triangles that can be obtained from these points }={ }^{12} C_3-{ }^4 C_3 \\
& =220-\frac{4 !}{3 ! \times 1 !} \\
& =220-\frac{4 \times 3 !}{3 !} \\
& =220-4 \\
& =216
\end{aligned}
$
View full question & answer→Question 241 Mark
Find the number of triangles formed by joining 12 points if: no three points are collinear
AnswerThere are 12 points on the plane: When no three of them are collinear:
Since a triangle can be drawn by joining any three non-collinear points.
$
\begin{aligned}
& \therefore \text { Number of triangles that can be obtained from these points }={ }^{12} C_3 \\
& =\frac{12 !}{3 ! ! !} \\
& =\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !} \\
& =220
\end{aligned}
$
View full question & answer→Question 251 Mark
Find the value of: ${ }^{20} \mathrm{C}_{16}-{ }^{19} \mathrm{C}_{16}$
Answer$
\begin{aligned}
{ }^{20} \mathrm{C}_{16}-{ }^{19} \mathrm{C}_{16} & =\frac{20 !}{16 ! 4 !}-\frac{19 !}{16 ! 3 !} \\
& =\frac{20 \times 19 !}{16 ! \times 4 \times 3 !}-\frac{19 !}{16 ! 3 !} \\
& =\frac{19 !}{3 ! 16 !}\left[\frac{20}{4}-1\right] \\
& =\frac{19 !}{3 ! 16 !}(4) \\
& =\frac{19 !}{3 !(16)(15 !)} 4 \\
& =\frac{19 !}{4(3 !)(15 !)} \\
& =\frac{19 !}{4 ! 15 !} \\
& ={ }^{19} \mathrm{C}_{15} \text { or }{ }^{19} \mathrm{C}_4=3876
\end{aligned}
$
View full question & answer→Question 261 Mark
Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if: four points are collinear
AnswerThere are 10 points on a plane.: When 4 of them arc collinear:
$\therefore$ Number of lines passing through these points if 4 points are collinear
$
\begin{aligned}
& ={ }^{10} C_2-{ }^4 C_2+1 \\
& =45-\frac{4 !}{2 ! 2 !}+1 \\
& =45-\frac{4 \times 3 \times 2 !}{2 \times 2 !}+1 \\
& =45-6+1 \\
& =40
\end{aligned}
$
View full question & answer→Question 271 Mark
Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if: no three points are collinear
AnswerThere are 10 points on a plane.: No three of them are collinear:
Since a line is obtained by joining 2 points,
number of lines passing through these points if no three points are collinear $={ }^{10} \mathrm{C}_2$
$
\begin{aligned}
& =\frac{10 !}{2 ! 8 !} \\
& =\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !} \\
& =5 \times 9 \\
& =45
\end{aligned}
$
View full question & answer→Question 281 Mark
Find the value of: ${ }^{15} C_4+{ }^{15} C_5$
Answer$
\begin{aligned}
{ }^{15} \mathrm{C}_4+{ }^{15} \mathrm{C}_5 & ={ }^{15} \mathrm{C}_5+{ }^{15} \mathrm{C}_4={ }^{15} \mathrm{C}_5+{ }^{15} \mathrm{C}_5-1 \\
& ={ }^{16} \mathrm{C}_5 \quad \ldots\left[\because \because{ }^n \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}\right] \\
& =4368
\end{aligned}
$
View full question & answer→Question 291 Mark
There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
AnswerThere are 20 lines such that no two of them are parallel and no three of them are concurrent.
Since no two lines are parallel
$\therefore$ they intersect at a point
$\therefore$ Number of points of intersection if no two lines are parallel and no three lines are
concurrent $={ }^{20} \mathrm{C}_2$
$
\begin{aligned}
& =\frac{20 !}{2 ! 18 !} \\
& =\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !} \\
& =190
\end{aligned}
$
View full question & answer→Question 301 Mark
Find the value of: ${ }^{80} \mathrm{C}_2$
Answer$
\begin{aligned}
&{ }^{80} \mathrm{C}_2=\frac{80 !}{2 !(80-2) !}=\frac{80 !}{2 ! 78 !} \\
&=\frac{80 \times 79 \times 78 !}{2 \times 78 !} \\
& \\
&=40 \times 79=3160
\end{aligned}
$
View full question & answer→Question 311 Mark
Find the number of diagonals of an $n$-sided polygon. In particular, find the number of diagonals when: $\mathrm{n}=12$
Answer$
\begin{aligned}
\mathrm{n}=12,{ }^{\mathrm{n}} \mathrm{C}_2-\mathrm{n} & ={ }^{12} \mathrm{C}_2-12 \\
& =\frac{12 \times 11}{1 \times 2}-12 \\
& =66-12=54
\end{aligned}
$
View full question & answer→Question 321 Mark
Find the number of diagonals of an $n$-sided polygon. In particular, find the number of diagonals when: $n=15$
Answer$
\begin{aligned}
\mathrm{n}=15,{ }^{\mathrm{n}} \mathrm{C}_2-\mathrm{n} & ={ }^{15} \mathrm{C}_2-15 \\
& =\frac{15 \times 14}{1 \times 2}-15 \\
& =105-15=90
\end{aligned}
$
View full question & answer→Question 331 Mark
Find the number of diagonals of an $n$-sided polygon. In particular, find the number of diagonals when: $\mathrm{n}=10$
AnswerIn $n$-sided polygon, there are ' $n$ ' points and ' $n$ ' sides. .
$\therefore$ Through ' $n$ ' points we can draw ${ }^n C_2$ lines including sides.
$\therefore$ Number of diagonals in $\mathrm{n}$ sided polygon $={ }^{\mathrm{n}} \mathrm{C}_2-\mathrm{n}(\therefore \mathrm{n}=$ number of sides)
i.
$
\begin{aligned}
& \mathrm{n}=10, \\
&{ }^{\mathrm{n}} \mathrm{C}_2-\mathrm{n}={ }^{10} \mathrm{C}_2-10 \\
&=\frac{10 \times 9}{1 \times 2}-10 \\
&=45-10=35
\end{aligned}
$
View full question & answer→Question 341 Mark
Find the value of: ${ }^{15} \mathrm{C}_4$
Answer$
\begin{aligned}
{ }^{15} \mathrm{C}_4 & =\frac{15 !}{4 !(15-4) !}=\frac{15 !}{4 ! 11 !} \\
& =\frac{15 \times 14 \times 13 \times 12 \times 11 !}{4 \times 3 \times 2 \times 1 \times 11 !} \\
& =\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} \\
& =1365
\end{aligned}
$
View full question & answer→Question 351 Mark
If 20 points are marked on a circle, how many chords can be drawn?
AnswerTo draw a chord we need to join two points on the circle. There are 20 points on a circle.
$
\begin{aligned}
& \therefore \text { Total number of chords possible from these points }={ }^{20} \mathrm{C}_2 \\
& =\frac{20 !}{2 ! 18 !} \\
& =\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !} \\
& =190
\end{aligned}
$
View full question & answer→Question 361 Mark
Four objects in a set of ten objects are alike. Find the number of ways of arranging them in a circular order.
AnswerTen things can be arranged in a circular order of which 4 are alike in $\frac{9 !}{4 !}$ ways.
$\therefore$ Required total number of arrangements $=\frac{9 !}{4 !}$
View full question & answer→Question 371 Mark
Find the number of seating arrangements for 3 men and 3 women to sit around a table so that exactly two women are together.
Question 381 Mark
Eight men and six women sit around a table. How many sitting arrangements will have no two women together?
Answer8 men can be seated around a table in $(8-1) !=7 !$ ways.
There are 8 gaps created by 8 men's seats.
$\therefore 6$ Women can be seated in 8 gaps in ${ }^8 \mathrm{P}_6$ ways
$\therefore$ Total number of arrangements so that no two women are together $=7 ! \times{ }^8 \mathrm{P}_6$
View full question & answer→Question 391 Mark
Five men, two women, and a child sit around a table. Find the number of arrangements where the child is seated: between two men.
AnswerTwo men out of 5 men can sit on either side of the child in 5P2 ways.
Let us take two men and a child as one unit.
Now these are to be arranged with the remaining 3 men and 2 women i.e., a total of 6 events $(3+2+1)$ is to be arranged around a round table which can be done in $(6-1) !=5$ ! ways.
$\therefore$ The total number of arrangements, if the child is seated between two men $={ }^5 P_2 \times$ 5 !
View full question & answer→Question 401 Mark
Five men, two women, and a child sit around a table. Find the number of arrangements where the child is seated: between the two women.
Answer5 men, 2 women, and a child sit around a table
When a child is seated between two women
$\therefore$ The two women can be seated on either side of the child in 2! ways.
Let us consider these 3 (two women and a child) as one unit.
Also, these 3 are to be seated with 5 men, (i.e. a total of 6 units) which can be done in $(6-1) !=5$ ! ways.
$\therefore$ The total number of arrangements if the child is seated between two women $=5 ! \times$
2 !
View full question & answer→Question 411 Mark
A committee of 20 members sits around a table. Find the number of arrangements that have the president and the vice president together.
AnswerA committee of 20 members sits around a table.
But, President and Vice-president sit together.
Let us consider President and Vice-president as one unit.
They can be arranged among themselves in 2! ways.
Now, this unit with the other 18 members of the committee is to be arranged around a table, which can be done in $(19-1) !=18$ ! ways.
$\therefore$ The total number of arrangements possible if President and Vice-president sit together $=18 ! \times 2$ !
View full question & answer→Question 421 Mark
Delegates from 24 countries participate in a round table discussion. Find the number of seating arrangements where two specified delegates are: never together.
AnswerWhen 2 specified delegates are never together then, the other 22 delegates can participate in a round table discussion in $(22-1) !=21$ ! ways.
$\therefore$ There are 22 places of which any 2 places can be filled by those 2 delegates who are never together.
$\therefore$ Two specified delegates can be arranged in ${ }^{22} \mathrm{P}_2$ ways.
$\therefore$ Total number of arrangements if two specified delegates are never together $={ }^{22} \mathrm{P}_2$
$\times 21$ !
$=\frac{22 !}{(22-2) !} \times 21 !$
$=\frac{22 !}{20 !} \times 21 !$
$=22 \times 21 \times 21$ !
$=21 \times 22 \times 21$ !
$=21 \times 22$ !
View full question & answer→Question 431 Mark
Delegates from 24 countries participate in a round table discussion. Find the number of seating arrangements where two specified delegates are: always together.
AnswerDelegates of 24 countries are to participate in a round table discussion such that two specified delegates are always together.
Let us consider these 2 delegates as one unit.
They can be arranged among themselves in 2! ways.
Also, these two delegates are to be seated with 22 other delegates (i.e. total 23) which can be done in $(23-1) !=22$ ! ways.
$\therefore$ The total number of arrangements if two specified delegates are always together $=$ $22 ! \times 2$ !
View full question & answer→Question 441 Mark
In how many different ways can 8 friends sit around a table?
AnswerWe know that ' $n$ ' persons can sit around a table in ( $n-1)$ ! ways
$\therefore 8$ friends can sit around a table in 7 ! ways
$
\begin{aligned}
& =7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\
& =5040 \text { ways. }
\end{aligned}
$
$\therefore 8$ friends can sit around a table in 5040 ways.
View full question & answer→Question 451 Mark
Find the number of arrangements of letters in the word CONSTITUTION that begin and end with $\mathrm{N}$.
AnswerThere are 12 letters in the word CONSTITUTION, in which 'O', 'N', T repeat two times each, ' $T$ ' repeats 3 times.
The arrangement starts and ends with ' $N$ ', 10 letters other than $\mathrm{N}$ can be arranged between two $\mathrm{N}$, in which ' $\mathrm{O}$ ' and ' $\mathrm{I}$ ' repeat twice each and ' $\mathrm{T}$ ' repeats 3 times.
$\therefore$ Total number of arrangements with the letter $\mathrm{N}$ at the beginning and at the end $=$ $\frac{10 !}{2 ! 2 ! 3 !}$
View full question & answer→Question 461 Mark
Find the number of arrangements of letters in the word MUMBAl so that the letter $B$ is always next to $\mathrm{A}$.
AnswerThere are 6 letters in the word MUMBAI.
These letters are to be arranged in such a way that ' $B$ ' is always next to ' $A$ '.
Let us consider $\mathrm{AB}$ as one unit. This unit with other 4 letters in which ' $\mathrm{M}$ ' repeats twice, is to be arranged.
$\therefore$ Total number of arrangements when $B$ is always next to $A=\frac{5 !}{2 !}$
$
\begin{aligned}
& =\frac{5 \times 4 \times 3 \times 2 !}{2 !} \\
& =60
\end{aligned}
$
View full question & answer→Question 471 Mark
A bag has 5 red, 4 blue, and 4 green marbles. If all are drawn one by one and their colours are recorded, how many different arrangements can be found?
AnswerThere is a total of 13 marbles in a bag.
Out of these 5 are Red, 4 Blue, and 4 are Green marbles.
All balls of the same colour are taken to be identical.
$\therefore$ Required number of arrangements $=\frac{13 !}{5 ! 4 ! 4 !}$
View full question & answer→Question 481 Mark
A coin is tossed 8 times. In how many ways can we obtain: at least 6 heads?
AnswerA coin is tossed 8 times. All heads are identical and all tails are identical.: When at least 6 heads are to be obtained
$\therefore$ Outcome can be ( 6 heads and 2 tails) or ( 7 heads and 1 tail) or ( 8 heads)
$\therefore$ Number of ways in which it can be obtained $=\frac{8 !}{6 ! 2 !}+\frac{8 !}{7 ! 1 !}+\frac{8 !}{8 !}$
$=\frac{8 \times 7}{2}+8+1$
$=28+8+1$
$=37$
$\therefore$ In 37 different ways we can obtain at least 6 heads.
View full question & answer→Question 491 Mark
A coin is tossed 8 times. In how many ways can we obtain: 4 heads and 4 tails?
AnswerA coin is tossed 8 times. All heads are identical and all tails are identical.: We can obtain 4 heads and 4 tails in $\frac{8 !}{4 ! 4 !}$
$
\begin{aligned}
& =\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2} \\
& =70 \text { ways }
\end{aligned}
$
$\therefore$ In 70 different ways we can obtain 4 heads and 4 tails.
View full question & answer→Question 501 Mark
You have 2 identical books on English, 3 identical books on Hindi and 4 identical books on Mathematics. Find the number of distinct ways of arranging them on a shelf.
AnswerThere are total 9 books to be arranged on a shelf.
Out of these 9 books, 2 books on English, 3 books on Hindi and 4 books on Mathematics are identical.
$
\begin{aligned}
& \therefore \text { Total number of arrangements }=\frac{9 !}{2 ! 3 ! 4 !} \\
& =\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{2 \times 3 \times 2 \times 4 !} \\
& =9 \times 4 \times 7 \times 5 \\
& =1260
\end{aligned}
$
$\therefore$ In 1260 distinct ways the books can be arranged on a shelf.
View full question & answer→Question 511 Mark
Find the number of permutations of letters in each of the following words: COMBINE
AnswerThere are 7 distinct letters in the word COMBINE which can be arranged among themselves in $7 !=5040$ ways
View full question & answer→Question 521 Mark
Find the number of permutations of letters in each of the following words : REPRESENT
AnswerThere are 9 letters in the word REPRESENT in which ' $E$ ' repeats 3 times and ' $R$ ' repeats 2 times.
$\therefore$ Number of permutations of the letters of the word REPRESENT $=\frac{9 !}{3 ! 2 !}$
$
=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{2}
$
$
=30240
$
View full question & answer→Question 531 Mark
Find the number of distinct numbers formed using the digits $3,4,5,6,7,8,9$, so that odd positions are occupied by odd digits.
AnswerA number is to be formed with digits $3,4,5,6,7,8,9$ such that odd digits always occupy the odd places.
There are 4 odd digits i.e. 3, 5, 7, 9 .
They can be arranged at 4 odd places among themselves in 4 ! ways $=24$ ways
3 even places of the number are occupied by even digits (i.e. $4,6,8$ ).
$\therefore$ They can be arranged in 3 ! ways $=6$ ways
$\therefore$ Total number of arrangements $=24 \times 6=144$
$\therefore 144$ numbers can be formed so that odd digits always occupy the odd positions.
View full question & answer→Question 541 Mark
Find the number of permutations of letters in each of the following words: SHANTARAM
AnswerThere are 9 letters in the word SHANTARAM in which 'A' repeats 3 times.
$\therefore$ Number of permutations of the letters of the word SHANTARAM $=\frac{9}{3 !}$
$
\begin{aligned}
& =9 \times 8 \times 7 \times 6 \times 5 \times 4 \\
& =60480
\end{aligned}
$
View full question & answer→Question 551 Mark
Find the number of permutations of letters in each of the following words : DIVYA
Answer There are 5 letters in the word DIVYA which can be arranged in 5 ! Way $=120$ ways
View full question & answer→Question 561 Mark
Find the number of arrangements of the letters in the word BERMUDA so that consonants and vowels are in the same relative positions.
AnswerThere are 7 letters in the word "BERMUDA" out of which 3 are vowels and 4 are consonants.
If relative positions of consonants and vowels are not changed.
3 vowels can be arranged among themselves in ${ }^3 P_3$ i.e., 3 ! ways.
4 consonants can be arranged among themselves in ${ }^4 P_4$ i.e., 4 ! ways.
$\therefore$ total no. of arrangements possible if relative positions of vowels and consonants are not changed $=3 ! \times 4$ !
$=6 \times 24$
$=144$
View full question & answer→Question 571 Mark
How many 4 letter words can be formed using letters in the word MADHURI if: letters cannot be repeated?
AnswerWhen repetition of letters is not allowed, the number of 4-letter words formed from the letters of the word MADHURI is ${ }^7 \mathrm{P}_4=\frac{7 !}{(7-4) !}=\frac{7 \times 6 \times 5 \times 4 \times 3 !}{3 !}=840$
$\therefore 840$ four-letter words can be formed when repetition of letters is not allowed.
View full question & answer→Question 581 Mark
How many 4 letter words can be formed using letters in the word MADHURI if: letters can be repeated?
AnswerA 4 letter word is to be formed from the letters of the word MADHURI and repetition of letters is allowed.
$\therefore 1$ st letter can be filled in 7 ways.
2nd letter can be filled in 7 ways.
3 rd letter can be filled in 7 ways.
4 th letter can be filled in 7 ways.
$\therefore$ Total no. of ways a 4-letter word can be formed $=7 \times 7 \times 7 \times 7=2401$
$\therefore 2401$ four-lettered words can be formed when repetition of letters is allowed.
View full question & answer→Question 591 Mark
Show that $(n+1){ }^n P_r=(n-r+1)^{(n+1)} P_r$.
Answer$
\begin{aligned}
& \text { L.H.S. }=(n+1){ }^n P_r=(n+1) \frac{n !}{(n-r) !}=\frac{(n+1) !}{(n-r) !} \\
& \begin{aligned}
& \text { R.H.S. }=(n-r+1){ }^{(n+1)} P_r=(n-r+1) \frac{(n+1) !}{(n-r+1) !} \\
& \\
&=\frac{(n-r+1)(n+1) !}{(n-r+1)(n-r) !} \\
&=\frac{(n+1) !}{(n-r) !}
\end{aligned} \\
& \therefore \quad \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$
View full question & answer→Question 601 Mark
Find the number of arranging 11 distinct objects taken 4 at a time so that a specified object: never occurs
Answer When one specified object does not occur then 4 things are to be arranged from the remaining 10 things, which can be done in ${ }^{10} \mathrm{P}_4$ ways
$
=10 \times 9 \times 8 \times 7 \text { ways }
$
$
=5040 \text { ways }
$
$\therefore$ There are 5040 permutations of 11 distinct objects, taken 4 at a time, in which one specified object never occurs.
View full question & answer→Question 611 Mark
Find the number of arranging 11 distinct objects taken 4 at a time so that a specified object: always occurs
AnswerThe number of permutations of $n$ distinct objects, taken $r$ at a time, when one specified object will always occur is $r \times{ }^{(n-1)} P_{(r-1)}$
Here, $r=4, n=11$
$\therefore$ The number of permutations of 4 out of 11 objects when a specified object occurs.
$
\begin{aligned}
& =4 \times{ }^{(11-1)} P_{(4-1)} \\
& =4 \times{ }^{10} P_3 \\
& =4 \times \frac{10 !}{(10-3) !} \\
& =4 \times \frac{10 !}{7 !} \\
& =4 \times \frac{10 \times 9 \times 8 \times 7 !}{7 !} \\
& =2880
\end{aligned}
$
$\therefore$ There are 2880 permutations of 11 distinct objects, taken 4 at a time, in which one specified object always occurs.
View full question & answer→Question 621 Mark
Find the number of ways in which 5 letters can be posted in 3 post boxes if any number of letters can be posted in a post box.
AnswerThere are 5 letters and 3 post boxes and any number of letters can be posted in all three post boxes.
$\therefore$ Each letter can be posted in 3 ways.
$\therefore$ Total number of ways in which 5 letters can be posted $=3 \times 3 \times 3 \times 3 \times 3=243$
View full question & answer→Question 631 Mark
Find the number of 6 -digit numbers using the digits 3, 4, 5, 6, 7, 8 without repetition.
How many of these numbers are: not divisible by 5 ?
AnswerA number of 6 different digits is to be formed from the digits $3,4,5,6,7,8$ which can be done in ${ }^6 \mathrm{P}_6$ i.e., $6 !=720$ ways: If the number is not divisible by 5 , then
Unit's place can be any digit from $3,4,6,7,8$ which can be selected in 5 ways.
Other 5 digits can be arranged in ${ }^5 \mathrm{P}_5$ i.e., 5 ! ways
$\therefore$ The total number of ways in which numbers not divisible by 5 can be formed $=5 \times$
5 !
$
\begin{aligned}
& =5 \times 120 \\
& =600
\end{aligned}
$
View full question & answer→Question 641 Mark
Find the number of 6 -digit numbers using the digits 3, 4, 5, 6, 7, 8 without repetition.
How many of these numbers are: divisible by 5 ?
AnswerA number of 6 different digits is to be formed from the digits $3,4,5,6,7,8$ which can be done in ${ }^6 \mathrm{P}_6$ i.e., $6 !=720$ ways: If the number is divisible by 5 , then
The unit's place digit must be 5 , and hence unit's place can be filled in 1 way Other 5 digits can be arranged among themselves in ${ }^5 P_5$ i.e., 5 ! ways
$\therefore$ Total number of ways in which numbers divisible by 5 can be formed $=1 \times 5$ ! = 120
View full question & answer→Question 651 Mark
How many numbers can be formed using the digits $0,1,2,3,4,5$ without repetition so that the resulting numbers are between 100 and 1000 ?
AnswerA number between 100 and 1000 is a 3 digit number and is to be formed from the digits $0,1,2,3,4,5$, without repetition of digits.
$\therefore 100$ 's place digit must be a non-zero number which can be filled in 5 ways.
10 's place digits can be filled in 5 ways.
Unit's place digit can be filled in 4 ways.
$\therefore$ total number of ways the number can be formed $=5 \times 5 \times 4=100$
$\therefore 100$ numbers between 100 and 1000 can be formed.
View full question & answer→Question 661 Mark
Find the number of 4-digit numbers that can be formed using the digits $1,2,4,5,6$, 8 if: digits cannot be repeated
AnswerA 4 different digit number is to be made from the digits $1,2,4,5,6,8$ without repetition of digits.
$\therefore 4$ different digits are to be arranged from 6 given digits which can be done in ${ }^6 \mathrm{P}_4$ $=\frac{6 !}{(6-4) !}$
$=\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}$
$=360$ ways
$\therefore 360$ four-digit numbers can be formed, if repetition of digits is not allowed.
View full question & answer→Question 671 Mark
Find the number of 4-digit numbers that can be formed using the digits $1,2,4,5,6$, 8 if: digits can be repeated
AnswerA 4 digit number is to be made from the digits $1,2,4,5,6,8$ such that digits can be repeated.
$\therefore$ The unit's place digit can be filled in 6 ways.
10 's place digit can be filled in 6 ways.
100 's place digit can be filled in 6 ways.
1000 's place digit can be filled in 6 ways.
$\therefore$ total number of numbers $=6 \times 6 \times 6 \times 6=6^4=1296$
$\therefore 1296$ four-digit numbers can be formed if repetition of digits is allowed
View full question & answer→Question 681 Mark
Find n, if: $\frac{(15-n) !}{(13-n) !}=12$
Answer$
\begin{aligned}
& \frac{(15-n) !}{(13-n) !}=12 \\
& \therefore \frac{(15-\mathbf{n})(14-\mathbf{n})(13-\mathbf{n}) !}{(13-\mathbf{n}) !}=12 \\
& \therefore(15-\mathrm{n})(14-\mathrm{n})=4 \times 3
\end{aligned}
$
Comparing on both sides, we get
$
\begin{aligned}
& 15-n=4 \\
& \therefore n=11
\end{aligned}
$
View full question & answer→Question 691 Mark
Find $n$, if: $\frac{(17-n) !}{(14-n) !}=5$ !
Answer$
\begin{aligned}
& \frac{(17-n) !}{(14-n) !}=5 ! \\
& \therefore \frac{(17-n)(16-n)(15-n)(14-n) !}{(14-n) !}=5 \times 4 \times 3 \times 2 \times 1 \\
& \therefore(17-\mathrm{n})(16-\mathrm{n})(15-\mathrm{n})=6 \times 5 \times 4
\end{aligned}
$
Comparing on both sides, we get
$
\begin{aligned}
& 17-n=6 \\
& \therefore \mathrm{n}=11
\end{aligned}
$
View full question & answer→Question 701 Mark
Find n, if : $(n+3) !=110 \times(n+1)$ !
Answer$
\begin{aligned}
& (n+3) !=110 \times(n+1) ! \\
& \therefore(n+3)(n+2)(n+1) !=110(n+1) ! \\
& \therefore(n+3)(n+2)=(11)(10)
\end{aligned}
$
Comparing on both sides, we get
$
\begin{aligned}
& n+3=11 \\
& \therefore n=8
\end{aligned}
$
View full question & answer→Question 711 Mark
Find $n$, iF : $(n+1) !=42 \times(n-1) !$
Answer$
\begin{aligned}
& (n+1) !=42(n-1) ! \\
& \therefore(n+1) n(n-1) !=42(n-1) ! \\
& \therefore n^2+n=42 \\
& \therefore n(n+1)=6 \times 7
\end{aligned}
$
Comparing on both sides, we get
$
\therefore \mathrm{n}=6
$
View full question & answer→Question 721 Mark
Find n, if : $\frac{1}{n !}=\frac{1}{4 !}-\frac{4}{5 !}$
Answer$
\begin{array}{ll}
& \frac{1}{n !}=\frac{1}{4 !}-\frac{4}{5 !} \\
\therefore & \frac{1}{n !}=\frac{1}{4 !}-\frac{4}{5 !} \\
\therefore & \frac{1}{n !}=\frac{5}{5 \times 4 !}-\frac{4}{5 !} \\
\therefore & \frac{1}{n !}=\frac{5}{5 !}-\frac{4}{5 !} \\
\therefore & \frac{1}{n !}=\frac{1}{5 !} \\
\therefore & n !=5 ! \\
\therefore & n=5
\end{array}
$
View full question & answer→Question 731 Mark
Find n, if : $\frac{n}{6 !}=\frac{4}{8 !}+\frac{3}{6 !}$
Answer$
\begin{array}{ll}
& \frac{n}{6 !}=\frac{4}{8 !}+\frac{3}{6 !} \\
\therefore & \frac{n}{6 !}-\frac{3}{6 !}=\frac{4}{8 !} \\
\therefore & \frac{n-3}{6 !}=\frac{4}{8 \times 7 \times 6 !} \\
\therefore & n-3=\frac{4}{8 \times 7} \\
\therefore & n-3=\frac{1}{14} \\
\therefore & n=\frac{1}{14}+3=\frac{43}{14}
\end{array}
$
View full question & answer→Question 741 Mark
Find $n$, iF : $\frac{n}{8 !}=\frac{3}{6 !}+\frac{1}{4 !}$
Answer$
\begin{array}{ll}
& \frac{\mathrm{n}}{8 !}=\frac{3}{6 !}+\frac{1}{4 !} \\
\therefore \quad & \frac{\mathrm{n}}{8 !}=\frac{3}{6 !}+\frac{6 \times 5}{6 \times 5 \times 4 !} \\
\therefore \quad & \frac{\mathrm{n}}{8 !}=\frac{3}{6 !}+\frac{30}{6 !} \\
\therefore \quad & \frac{\mathrm{n}}{8 \times 7 \times 6 !}=\frac{33}{6 !} \\
\therefore \quad & \frac{\mathrm{n}}{56}=33 \\
\therefore \quad & \mathrm{n}=56 \times 33=1848
\end{array}
$
View full question & answer→Question 751 Mark
Evaluate: $\frac{n !}{r !(n-r) !}$ for : $n=12, r=12$
Answer$
\begin{aligned}
& n=12, r=12 \\
& \begin{aligned}
& \therefore \quad \frac{n !}{r !(n-r) !}=\frac{12 !}{12 !(12-12) !} \\
&=\frac{12 !}{12 ! 0 !} \\
&=1 \\
& \quad \ldots[\because 0 !=1]
\end{aligned}
\end{aligned}
$
View full question & answer→Question 761 Mark
Evaluate: $\frac{n !}{r !(n-r) !}$ for : $\mathrm{n}=8, \mathrm{r}=6$
Answer$
\begin{aligned}
& n=8, r=6 \\
& \therefore \quad \frac{n !}{r !(n-r) !}=\frac{8 !}{6 !(8-6) !} \\
& =\frac{8 \times 7 \times 6 !}{6 ! 2 !} \\
& =\frac{8 \times 7}{2 !} \\
& =\frac{8 \times 7}{1 \times 2}=28 \\
&
\end{aligned}
$
View full question & answer→Question 771 Mark
Write in terms of factorials: $5 \times 10 \times 15 \times 20 \times 25$
Answer$
\begin{aligned}
& 5 \times 10 \times 15 \times 20 \times 25 \\
& =(5 \times 1) \times(5 \times 2) \times(5 \times 3) \times(5 \times 4) \times(5 \times 5) \\
& =\left(5^5\right)(5 \times 4 \times 3 \times 2 \times 1) \\
& =\left(5^5\right)(5 !)
\end{aligned}
$
View full question & answer→Question 781 Mark
Write in terms of factorials: $6 \times 7 \times 8 \times 9$
Answer$
\begin{aligned}
& 6 \times 7 \times 8 \times 9 \\
& =9 \times 8 \times 7 \times 6
\end{aligned}
$
Multiplying and dividing by 5 !, we get
$
\begin{aligned}
& =\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 !} \\
& =\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 !}=\frac{9 !}{5 !}
\end{aligned}
$
View full question & answer→Question 791 Mark
Write in terms of factorials: $3 \times 6 \times 9 \times 12 \times 15$
Answer$
\begin{aligned}
& 3 \times 6 \times 9 \times 12 \times 15 \\
& =3 \times(3 \times 2) \times(3 \times 3) \times(3 \times 4) \times(3 \times 5) \\
& =\left(3^5\right)(5 \times 4 \times 3 \times 2 \times 1) \\
& =3^5(5 !)
\end{aligned}
$
View full question & answer→Question 801 Mark
Write in terms of factorials : $5 \times 6 \times 7 \times 8 \times 9 \times 10$
Answer$
\begin{aligned}
& 5 \times 6 \times 7 \times 8 \times 9 \times 10 \\
& =10 \times 9 \times 8 \times 7 \times 6 \times 5
\end{aligned}
$
Multiplying and dividing by 4 !, we get
$
\begin{aligned}
& =\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \text { MaharashtraBoardSolutions.in }} \\
& =\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 !}=\frac{10 !}{4 !}
\end{aligned}
$
View full question & answer→Question 811 Mark
Compute : $\frac{8 !}{(6-4) !}$
Answer$
\frac{8 !}{(6-4) !}=\frac{8 !}{2 !}=\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 !}{2 !}=20160
$
View full question & answer→Question 821 Mark
Compute:$\frac{8 !}{6 !-4 !}$
Answer$
\begin{aligned}
\frac{8 !}{6 !-4 !} & =\frac{8 \times 7 \times 6 \times 5 \times 4 !}{6 \times 5 \times 4 !-4 !} \\
& =\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 !(6 \times 5-1)}=\frac{1680}{29}
\end{aligned}
$
View full question & answer→Question 831 Mark
Compute:$\frac{6 !-4 !}{4 !}$
Answer$
\frac{6 !-4 !}{4 !}=\frac{6 \times 5 \times 4 !-4 !}{4 !}=\frac{4 !(6 \times 5-1)}{4 !}=29
$
View full question & answer→Question 841 Mark
Compute: $\frac{9 !}{3 ! 6 !}$
Answer$
\frac{9 !}{3 ! 6 !}=\frac{9 \times 8 \times 7 \times 6 !}{(3 \times 2 \times 1) \times 6 !}=84
$
View full question & answer→Question 851 Mark
Compute:3 ! $\times 2$ !
Answer$
\begin{aligned}
& 3 ! \times 2 ! \\
& =3 \times 2 \times 1 \times 2 \times 1 \\
& =12
\end{aligned}
$
View full question & answer→Question 861 Mark
Compute:$(3 \times 2)$ !
Answer$
\begin{aligned}
& (3 \times 2) ! \\
& =6 ! \\
& =6 \times 5 \times 4 \times 3 \times 2 \times 1 \\
& =720
\end{aligned}
$
View full question & answer→Question 871 Mark
Compute:$\left(\frac{12}{6}\right)$ !
Answer$
\begin{aligned}
& \left(\frac{12}{6}\right) ! \\
& =2 ! \\
& =2 \times 1 \\
& =2
\end{aligned}
$
View full question & answer→Question 881 Mark
Compute : $\frac{12 !}{6 !}$
Answer$
\begin{aligned}
& \frac{12 !}{6 !} \\
& =\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 !}{6 !} \\
& =12 \times 11 \times 10 \times 9 \times 8 \times 7 \\
& =665280
\end{aligned}
$
View full question & answer→Question 891 Mark
View full question & answer→Question 901 Mark
Find the value of: $\frac{5(26 !)+(27 !)}{4(27 !)-8(26 !)}$
Answer$
\begin{aligned}
\frac{5(26 !)+27 !}{4(27 !)-8(26 !)} & =\frac{5(26 !)+27(26 !)}{4(27 \times 26 !)-8(26 !)} \\
& =\frac{26 !(5+27)}{4(26 !)(27-2)} \\
& =\frac{32}{(4)(25)}=\frac{8}{25}
\end{aligned}
$
View full question & answer→Question 911 Mark
Find the value of: $\frac{8 !+5(4 !)}{4 !-12}$
Answer$
\begin{aligned}
\frac{8 !+5(4 !)}{4 !-12} & =\frac{8 !+5 !}{4 \times 3 \times 2-12} \\
& =\frac{8 \times 7 \times 6 \times 5 !+5 !}{4 \times 3 \times(2-1)} \\
& =\frac{5 !(8 \times 7 \times 6+1)}{4 \times 3} \\
& =\frac{5 \times 4 \times 3 \times 2 \times 1(336+1)}{4 \times 3} \\
& =5 \times 2 \times 337=3370
\end{aligned}
$
View full question & answer→Question 921 Mark
Answer8! – 6!
= 8 × 7 × 6! – 6!
= 6! (8 × 7 – 1)
= 6! (56 – 1)
= 6 × 5 × 4 × 3 × 2 × 1 × 55
= 39,600
View full question & answer→Question 931 Mark
Show that $\frac{9 !}{3 ! 6 !}+\frac{9 !}{4 ! 5 !}=\frac{10 !}{4 ! 6 !}$
Answer$
\begin{aligned}
\text { L.H.S. } & =\frac{9 !}{3 ! 6 !}+\frac{9 !}{4 ! 5 !}=\frac{9 !}{3 ! \times 6 \times 5 !}+\frac{9 !}{4 \times 3 ! \times 5 !} \\
& =\frac{9 !}{3 ! 5 !}\left[\frac{1}{6}+\frac{1}{4}\right] \\
& =\frac{9 !}{3 ! 5 !}\left[\frac{4+6}{6 \times 4}\right]=\frac{10 \times 9 !}{6 \times 5 ! \times 4 \times 3 !} \\
& =\frac{10 !}{6 ! 4 !}=\frac{10 !}{4 ! 6 !}=\text { R.H.S. }
\end{aligned}
$
View full question & answer→Question 941 Mark
Answer6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720
View full question & answer→Question 951 Mark
Answer8!
= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320
View full question & answer→Question 961 Mark
A school has three gates and four staircases from the first floor to the second floor. How many ways does a student have to go from outside the school to his classroom on the second floor?
AnswerA student can go inside the school from outside in 3 ways and from the first floor to the second floor in 4 ways.
∴ Number of ways to choose gates = 3
Number of ways to choose staircase = 4
∴ By using fundamental principle of multiplication,
number of ways in which a student has to go from outside the school to his classroom = 4 × 3 = 12
View full question & answer→Question 971 Mark
A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can he select a student if the monitor can be a boy or a girl?