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Maths (commerce) Solve the Following Question.(2 Marks)

Permutations and Combinations (p-2) · Maharashtra Board STD 11 Commerce / Arts (MSBSHSE - English Medium). 49 questions.

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Solve the Following Question.(2 Marks)

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49 questions · timed · auto-graded

Question 12 Marks
Thane is the 20th station from C.S.T. If a passenger can purchase a ticket from any station to any other station, how many different tickets must be available at the booking window?
Answer
Taking CST as the first station and Thane as 20th,
Let us name CST as $A_0$ next station as $A_1$ and so on, Thane is $A_{20}$
From station $A_0, 20$ different journeys are possible
From station $A_1, 20$ different journeys are possible.
From station $A_{20}, 20$ different journeys are possible.
Total number of different tickets of different journeys $=21 \times 20=420$
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Question 22 Marks
Find the number of ways of distributing $\mathrm{n}$ balls in $\mathrm{n}$ cells. What will be the number of ways if each cell must be occupied?
Answer
There are $\mathrm{n}$ balls and $\mathrm{n}$ cells
(i) Every ball can be put in any of the $n$ cells.
Number of distributions $=n \times n \times \ldots \ldots \times n=(n)^n$
(ii) For filling the first cell, $n$ balls are available.
The first cell is filled in $n$ ways.
The second cell is filled in $(n-1)$ ways
The third cell is filled in $(n-2)$ ways and so on.
the $n$th cell is tilled in one way.
Required number $=n(n-1)(n-2) \ldots . .1=n$ !
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Question 32 Marks
Find the number of words that can be formed by using all the letters in the word REMAIN. If these words are written in dictionary order, what will be the 40th word?
Answer
There are 6 letters $A, E, I, M, N, R$
The number of words starting with $A=5$ !
The number of words starting with $E=5$ !
The number of words starting with $\mathrm{I}=5$ !
The number of words starting with $M=5$ !
The number of words starting with $\mathrm{N}=5$ !
The number of words starting with $R=5$ !
Total number of words $=6 \times 5 !=720$
Number of words starting with $A E=4 !=24$
Number of words starting with AIE $=3 !=6$
Number of words starting with AIM $=3 !=6$
The number of words starting with AINE $=2$ !
Total words $=24+6+6+2=38$
39 th word is AINMER
40th word is AINMRE
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Question 42 Marks
Find the value of $\sum_{r=1}^4{ }^{21-r} C_4+{ }^{17} C_5$
Answer
$\sum_{\mathrm{r}=1}^4{ }^{(21-\mathrm{r})} \mathrm{C}_4+{ }^{17} \mathrm{C}_5=\left({ }^{20} \mathrm{C}_4+{ }^{19} \mathrm{C}_4+{ }^{18} \mathrm{C}_4+{ }^{17} \mathrm{C}_4\right)+{ }^{17} \mathrm{C}_5$
$=\left({ }^{20} \mathrm{C}_4+{ }^{19} \mathrm{C}_4+{ }^{18} \mathrm{C}_4+{ }^{18} \mathrm{C}_5-{ }^{17} \mathrm{C}_5\right)+{ }^{17} \mathrm{C}_5\text { ... }\left[{ }^n \mathrm{C}_{\mathrm{r}}+{ }^n \mathrm{C}_{\mathrm{r}-1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}\right]$
$={ }^{20} \mathrm{C}_4+{ }^{19} \mathrm{C}_4+\left({ }^{18} \mathrm{C}_4+{ }^{18} \mathrm{C}_5\right)$
$={ }^{20} \mathrm{C}_4+{ }^{19} \mathrm{C}_4+{ }^{19} \mathrm{C}_5$
$={ }^{20} \mathrm{C}_4+{ }^{20} \mathrm{C}_5$
$={ }^{21} \mathrm{C}_5$
$\sum_{\mathrm{r}=1}^4{ }^{(21-\mathrm{r})} \mathrm{C}_4+{ }^{17} \mathrm{C}_5={ }^{21} \mathrm{C}_5$
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Question 52 Marks
Find $r$ if ${ }^{11} C_4+{ }^{11} C_5+{ }^{12} C_6+{ }^{13} C_7={ }^{14} C_r$
Answer
$
\begin{aligned}
& { }^{11} \mathrm{C}_4+{ }^{11} \mathrm{C}_5+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_1 \\
& \therefore \quad\left({ }^{11} \mathrm{C}_4+{ }^{11} \mathrm{C}_5\right)+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_{\mathrm{T}} \\
& \cdots\left[{ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\right] \\
& \therefore \quad\left({ }^{12} \mathrm{C}_5+{ }^{12} \mathrm{C}_6\right)+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_r \\
& \therefore \quad\left({ }^{13} \mathrm{C}_6+{ }^{13} \mathrm{C}_7\right)={ }^{14} \mathrm{C}_{\mathrm{r}} \\
& \therefore \quad{ }^{14} \mathrm{C}_7={ }^{14} \mathrm{C}_{\mathrm{T}} \text { MaharashtraBoardSolutions.in } \\
& \text { If }{ }^{\mathrm{n}} \mathrm{C}_x={ }^{\mathrm{n}} \mathrm{C}_y \text {, then either } x=y \text { or } x=\mathrm{n}-y \\
& \therefore \quad \mathrm{r}=7 \text { or } \mathrm{r}=14-7=7 \\
&
\end{aligned}
$
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Question 62 Marks
Find $\mathrm{n}$ if ${ }^{21} \mathrm{C}_{6 \mathrm{n}}={ }^{21} C_{n^2+5}$
Answer
$
\begin{aligned}
& { }^{21} C_{6 n}={ }^{21} C_{n^2+5} \\
& \text { If }{ }^n C_x={ }^n C_y \text {, then either } x=y \text { or } x=n-y \\
& \therefore 6 n=n^2+5 \text { or } 6 n=21-\left(n^2+5\right) \\
& \therefore n^2-6 n+5=0 \text { or } 6 n=21-n^2-5 \\
& \therefore n^2-6 n+5=0 \text { or } n^2+6 n-16=0 \\
& \text { If } n^2-6 n+5=0 \text { then }(n-1)(n-5)=0 \\
& \therefore n=1 \text { or } n=5 \\
& \text { If } n=5 \text { then } n^2+5=30>21 \\
& \therefore n \neq 5 \\
& \therefore n=1 \\
& \text { If } n^2+6 n-16=0 \text { then }(n+8)(n-2)=0 \\
& n=-8 \text { or } n=2 \\
& n \neq-8 \\
& \therefore n=2
\end{aligned}
$
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Question 72 Marks
Five students are selected from 11. How many ways can these students be selected if: two specified students are not selected?
Answer
When 2 specified students are not included then 5 students can be selected from the remaining $(11-2)=9$ students
$\therefore$ Number of ways of selecting 5 students from 9 students $={ }^9 C_5$
$=\frac{9 !}{5 ! 4 !}$
$=\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1}$
$=126$
$\therefore$ Selection of students is done in 126 ways when 2 specified students are not selected.
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Question 82 Marks
Five students are selected from 11. How many ways can these students be selected if: two specified students are selected?
Answer
5 students are to be selected from 11 students
(i) When 2 specified students are included
then remaining 3 students can be selected from $(11-2)=9$ students.
$\therefore$ Number of ways of selecting 3 students from 9 students $={ }^9 C_3$
$
\begin{aligned}
& =\frac{9 !}{3 ! \times 6 !} \\
& =\frac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !} \\
& =84
\end{aligned}
$
$\therefore$ Selection of students is done in 126 ways when 2 specified students are not selected.
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Question 92 Marks
There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 players is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team. How many different teams can be formed?
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Question 102 Marks
A question paper has two sections. Section I has 5 questions and section II has 6 questions. A student must answer at least two questions from each section among 6 questions he answers. How many different choices does the student have in choosing questions?
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Question 112 Marks
A group consists of 9 men and 6 women. A team of 6 is to be selected. How many possible selections will have at least 3 women?
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Question 132 Marks
Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Answer
There are 6 boys and 4 girls.
A team of 3 boys and 2 girls is to be selected.
$\therefore 3$ boys can be selected from 6 boys in ${ }^6 \mathrm{C}_3$ ways.
2 girls can be selected from 4 girls in ${ }^4 C_2$ ways.
$\therefore$ Number of ways the team can be selected $={ }^6 \mathrm{C}_3 \times{ }^4 \mathrm{C}_2$
$=\frac{6 !}{3 ! 3 !} \times \frac{4 !}{2 ! 2 !}$
$=\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times \frac{4 \times 3 \times 2 !}{2 \times 2 !}$
$=20 \times 6$
$=120$
$\therefore$ The team of 3 boys and 2 girls can be selected in 120 ways.
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Question 142 Marks
Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 5 green balls and 7 blue balls so that 3 balls of every colour are drawn.
Answer
9 balls are to be selected from 6 red, 5 green, 7 blue balls such that the selection consists of 3 balls of each colour.
$\therefore 3$ red balls can be selected from 6 red balls in ${ }^6 \mathrm{C}_3$ ways.
3 green balls can be selected from 5 green balls in ${ }^5 C_3$ ways.
3 blue balls can be selected from 7 blue balls in ${ }^7 \mathrm{C}_3$ ways.
$\therefore$ Number of ways selection can be done if the selection consists of 3 balls of each colour
$
\begin{aligned}
& ={ }^6 \mathrm{C}_3 .{ }^5 \mathrm{C}_3 .{ }^7 \mathrm{C}_3 \\
& =\frac{6 !}{3 ! 3 !} \times \frac{5 !}{3 ! 2 !} \times \frac{7 !}{3 ! 4 !} \\
& =\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times \frac{5 \times 4 \times 3 !}{3 ! \times 2} \times \frac{7 \times 6 \times 5 \times 4 !}{3 \times 2 \times 1 \times 4 !} \\
& =20 \times 10 \times 35 \\
& =7000
\end{aligned}
$
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Question 152 Marks
Find $r$ if ${ }^{14} \mathrm{C}_{2 \mathrm{r}}:{ }^{10} \mathrm{C}_{2 \mathrm{r}-4}=143: 10$
Answer
$
\begin{aligned}
& { }^{14} \mathrm{C}_{2 \mathrm{r}}:{ }^{10} \mathrm{C}_{2 \mathrm{r}-4}=143: 10 \\
& \therefore \quad \frac{14 !}{2 r !(14-2 r) !} \div \frac{10 !}{(2 r-4) !(14-2 r) !}=\frac{143}{10} \\
& \therefore \quad \frac{14 !}{2 \mathrm{r} !(14-2 \mathrm{r}) !} \times \frac{(2 \mathrm{r}-4) !(14-2 \mathrm{r}) !}{10 !}=\frac{143}{10} \\
& \therefore \quad \frac{14 \times 13 \times 12 \times 11 \times 10 !}{2 r(2 r-1)(2 r-2)(2 r-3)(2 r-4) !(14-2 r) !} \\
& \times \frac{(2 r-4) !(14-2 r) !}{10 !}=\frac{143}{10} \\
& \\
& \therefore \frac{14 \times 13 \times 12 \times 11}{2 \mathrm{r}(2 \mathrm{r}-1) \times(2 \mathrm{r}-2)(2 \mathrm{r}-3)}=\frac{143}{10} \\
& \therefore 2 \mathrm{r}(2 \mathrm{r}-1)(2 \mathrm{r}-2)(2 \mathrm{r}-3)=14 \times 12 \times 10 \\
& \therefore 2 \mathrm{r}(2 \mathrm{r}-1)(2 \mathrm{r}-2)(2 \mathrm{r}-3)=8 \times 7 \times 6 \times 5 \\
& \text { Comparing on both sides, we get } \\
& \therefore r=4 \\
&
\end{aligned}
$
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Question 162 Marks
Find n if: ${ }^{2 n} C_3:{ }^n C_2=52: 3$
Answer
$
\begin{array}{ll}
& { }^{2 n} \mathrm{C}_3:{ }^{\mathrm{n}} \mathrm{C}_2=52: 3 \\
\therefore \quad & \frac{(2 \mathrm{n}) !}{3 !(2 \mathrm{n}-3) !} \div \frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}=\frac{52}{3} \\
\therefore \quad & \frac{(2 \mathrm{n}) !}{3 !(2 \mathrm{n}-3) !} \times \frac{2 !(\mathrm{n}-2) !}{\mathrm{n} !}=\frac{52}{3} \\
\therefore \quad & \frac{(2 \mathrm{n})(2 \mathrm{n}-1)(2 \mathrm{n}-2)(2 \mathrm{n}-3) !}{3 \times 2 !(2 \mathrm{n}-3) !} \times \frac{2 !(\mathrm{n}-2) !}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}=\frac{52}{3} \\
\therefore \quad & \frac{2 \mathrm{n}(2 \mathrm{n}-1) \cdot 2(\mathrm{n}-1)}{3} \times \frac{1}{\mathrm{n}(\mathrm{n}-1)}=\frac{52}{3} \\
\therefore & \frac{4(2 \mathrm{n}-1)}{3}=\frac{52}{3} \\
\therefore \quad & 2 \mathrm{n}-1=13 \\
\therefore \quad & 2 \mathrm{n}=14 \\
\therefore \quad & \mathrm{n}=7
\end{array}
$
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Question 172 Marks
A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Answer
Out of 8 consonants, 4 can be selected in ${ }^8 \mathrm{C}_4$
$
\begin{aligned}
& =\frac{8 !}{4 ! 4 !} \\
& =\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !} \\
& =70 \text { ways }
\end{aligned}
$
From 3 vowels, 2 can be selected in ${ }^3 \mathrm{C}_2$
$
\begin{aligned}
& =\frac{3 !}{2 ! 1 !} \\
& =\frac{3 \times 2 !}{2 !} \\
& =3 \text { ways }
\end{aligned}
$
Now, to form a word, these 6 letters (i.e., 4 consonants and 2 vowels) can be arranged in ${ }^6 \mathrm{P}_6$ i.e., 6 ! ways.
$
\begin{aligned}
& \therefore \text { Total number of words that can be formed }=70 \times 3 \times 6 ! \\
& =70 \times 3 \times 720 \\
& =151200 \\
& \therefore 151200 \text { words of } 4 \text { consonants and } 2 \text { vowels can be formed. }
\end{aligned}
$
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Question 182 Marks
Find the number of ways for 15 people to sit around the table so that no two arrangements have the same neighbours.
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Question 192 Marks
A party has 20 participants and a host. Find the number of distinct ways for the host to sit with them around a circular table. How many of these ways have two specified persons on either side of the host?
Answer
A party has 20 participants.
All of them and the host (i.e., 21 persons) can be seated at a circular table in $(21-1)$ ! $=20$ ! ways.
When two particular participants be seated on either side of the host.
Host takes chair in 1 way.
These 2 persons can sit on either side of host in 2! ways
Once host occupies his chair, it is not circular permutation any more.
Remaining 18 people occupy their chairs in 18 ! ways.
$\therefore$ Total number of arrangement possible if two particular participants be seated on either side of the host $=2 ! \times 18$ !
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Question 202 Marks
Fifteen persons sit around a table. Find the number of arrangements that have two specified persons not sitting side by side.
Answer
Since 2 particular persons can't be sitting side by side.
The other 13 persons can be arranged around the table in $(13-1) !=12$ !
13 people around a table create 13 gaps in which 2 people are to be seated
Number of arrangements of 2 people $={ }^{13} \mathrm{P}_2$
$\therefore$ The total number of arrangements in which two specified persons not sitting side
$
\begin{aligned}
& \text { by side }=12 ! \times{ }^{13} P_2 \\
& =12 ! \times 13 \times 12 \\
& =13 \times 12 ! \times 12 \\
& =12 \times 13 !
\end{aligned}
$
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Question 212 Marks
How many different words are formed if the letter $\mathrm{R}$ is used thrice and letters S and T are used twice each?
Answer
When ' $R$ ' is used thrice, ' $S$ ' is used twice and ' $T$ ' is used twice,
$\therefore$ Total number of letters available $=7$, of which ' $S$ ' and ' $T$ ' repeat 2 times each, ' $R$ ' repeats 3 times.
$\therefore$ Required number of arrangements $=\frac{7 !}{2 ! 2 ! 3 !}$
$=\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 1 \times 2 \times 1 \times 3 !}$
$=7 \times 6 \times 5$
$=210$
$\therefore 210$ different words can be formed with the letter $\mathrm{R}$ is used thrice and letters S and T are used twice each.
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Question 222 Marks
Find the number of different arrangements of letters in the word MAHARASHTRA. How many of these arrangements have: all vowels together?
Answer
There are 11 letters in the word MAHARASHTRA in which 'A' is repeated 4 times, ' $H$ ' repeated 2 times, and ' $R$ ' repeated 2 times.
$\therefore$ Total number of arrangements is $\frac{11 !}{4 ! 2 ! 2 !}$
$\therefore \frac{11 !}{4 ! 2 ! 2 !}$ different words can be formed from the letters of the word MAHARASHTRA.: When all vowels are together.
There are 4 vowels in the word MAHARASHTRA i.e., A, A, A, A
Let us consider these 4 vowels as one unit, they themselves can be arranged in $\frac{4 !}{4 !}=1$ way.
This unit is to be arranged with 7 other letters which can be done in 8 ! ways
$\therefore$ Total number of arrangements $=\frac{8 !}{2 ! 2 !}$
$\therefore \frac{8 !}{2 ! 2 !}$ different words can be formed if vowels are always together.
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Question 232 Marks
Find the number of different arrangements of letters in the word MAHARASHTRA. How many of these arrangements have: letters $M$ and T never together?
Answer
There are 11 letters in the word MAHARASHTRA in which 'A' is repeated 4 times, ' $H$ ' repeated 2 times, and ' $R$ ' repeated 2 times.
$\therefore$ Total number of arrangements is $\frac{11 !}{4 ! 2 ! 2 !}$
$\therefore \frac{11 !}{4 ! 2 ! 2 !}$ different words can be formed from the letters of the word MAHARASHTRA.: Other than $M$ and $T$. there are 9 letters in which A repeats 4 times, $H$ repeats twice, $R$ repeats twice
The number of arrangements of the a letter $=\frac{9 !}{4 ! 2 ! 2 !}$
These 9 letters create 10 gaps in which $M$ and $T$ are to be arranged
The number of arrangements of $\mathrm{M}$ and $\mathrm{T}={ }^{10} \mathrm{P}_2$
$\therefore$ Total number arrangement having $\mathrm{M}$ and $\mathrm{T}$ never together $=\frac{9 ! \times{ }^{10} \mathrm{P}_2}{4 ! 2 ! 2 !}$
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Question 242 Marks
Find the number of ways of arranging letters of the word MATHEMATICAL. How many of these arrangements have all vowels together?
Answer
There are 12 letters in the word MATHEMATICAL in which ' $M$ ' repeats 2 times, 'A' repeats 3 times, and ' $T$ ' repeats 2 times.
$\therefore$ Total number of arrangements $=\frac{12 !}{2 ! 3 ! 2 !}$
When all the vowels
i.e., 'A', 'A', 'A', ' $E$ ', $T$ are to be kept together
Number of arrangements of these vowels $=\frac{5 !}{3 !}$ ways.
Let us consider these vowels together as one unit.
This unit is to be arranged with 7 other letters in which ' $M$ ' and ' $T$ ' repeated 2 times each.
$\therefore$ Number of arrangements $=\frac{8 !}{2 ! 2 !}$
$\therefore$ Total number of arrangements $=\frac{8 ! \times 5 !}{2 ! 2 ! 3 !}$
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Question 252 Marks
Find the number of different ways of arranging letters in the word PLATOON if: consonants and vowels occupy alternate positions.
Answer
When consonants and vowels occupy alternate positions:
There are 4 consonants and 3 vowels in the word PLATOON.
$\therefore$ At odd places consonants occur and at even places vowels occur.
4 consonants can be arranged among themselves in 4 ! ways
3 vowels in which $\mathrm{O}$ occurs twice and $\mathrm{A}$ occurs once.
$\therefore$ They can be arranged in $\frac{3 !}{2 !}$ ways
$\therefore$ Required number of arrangements if the consonants and vowels occupy alternate
$
\begin{aligned}
& \text { positions }=4 ! \times \frac{3 !}{2 !} \\
& =4 \times 3 \times 2 \times \frac{3 \times 2 !}{2 !} \\
& =72
\end{aligned}
$
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Question 262 Marks
Find the number of different ways of arranging letters in the word PLATOON if: the two O's are never together.
Answer
When the two O's are never together:
Let us arrange the other 5 letters first, which can be done in $5 !=120$ ways.
The letters $\mathrm{P}, \mathrm{L}, \mathrm{A}, \mathrm{T}, \mathrm{N}$ create $6 \mathrm{gaps}$, in which O's are arranged.
$\therefore$ Two O's in 6 gaps can be arranged in $\frac{{ }^6 \mathrm{P}_2}{2 !}$ ways
$
\begin{aligned}
& =\frac{\frac{6 !}{(6-2) !}}{2 !} \text { ways } \\
& =\frac{6 \times 5 \times 4 !}{4 ! \times 2 \times 1} \text { ways } \\
& =3 \times 5 \text { ways } \\
& =15 \text { ways }
\end{aligned}
$
$\therefore$ Total number of arrangements if the two O's are never together $=120 \times 15=1800$
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Question 272 Marks
How many different 6-digit numbers can be formed using digits in the number 659942 ?How many of them are divisible by 2 ?
Answer
A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice.
$
\begin{aligned}
& \therefore \text { Total number of arrangements }=\frac{6 !}{2 !} \\
& =\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !} \\
& =360
\end{aligned}
$
$\therefore 360$ different 6 -digit numbers can be formed.
For a number to be divisible by 2 ,
Last digits can be selected in 3 ways
Remaining 5 digit in which, 9 appears twice are arranged in $\frac{5 !}{2 !}$ ways
$\therefore$ Total number of arrangements $=\frac{5 !}{2 !} \times 3=180$
$\therefore 180$ numbers are divisible by 2 .
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Question 282 Marks
How many distinct 5 digit numbers can be formed using the digits $3,2,3,2,4,5$.
Answer
5 digit numbers are to be formed from $2,3,2,3,4,5$.
Case I: Numbers formed from 2, 2, 3, 4, 5 OR 2, 3, 3, 4, 5
Number of such numbers $=\frac{5 !}{2 !} \times 2$
$
\begin{aligned}
& =5 ! \\
& =120
\end{aligned}
$
Case II: Numbers formed from 2, 2, 3, 3 and any one of 4 or 5
Number of such numbers $=\frac{5 !}{2 ! 2 !} \times 2=60$
Required number $=120+60=180$
$\therefore 180$ distinct 5 digit numbers can be formed using the digit $3,2,3,2,4,5$.
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Question 292 Marks
Find the number of ways letters of the word HISTORY can be arranged if: $\mathrm{Y}$ is next to $\mathrm{T}$.
Answer
There are 7 letters in the word HISTORY: When ' $Y$ ' is next to ' $T$ '
Let us take this (' $Y$ ' next to ' $T$ ') as one unit.
This unit with 5 other letters is to be arranged.
$\therefore$ The number of arrangements of 6 letters and one unit $={ }^6 \mathrm{P}_6=6$ !
Also ' $Y$ ' has to be always next to ' $T$ '.
So they can be arranged in 1 way.
$\therefore$ total number of arrangements possible when $Y$ is next to $T=6 ! \times 1=720$
$\therefore 720$ words can be formed if $Y$ is next to $T$.
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Question 302 Marks
Find the number of ways letters of the word HISTORY can be arranged if: $\mathrm{Y}$ and $\mathrm{T}$ are together
Answer
There are 7 letters in the word HISTORY: When ' $Y$ ' and ' $T$ ' are together.
Let us consider ' $Y$ ' and ' $T$ ' as one unit
This unit with the other 5 letters is to be arranged.
$\therefore$ The number of arrangements of one unit and 5 letters $={ }^6 \mathrm{P}_6=6$ !
Also, ' $Y$ ' and ' $T$ ' can be arranged among themselves in ${ }^2 P_2$ i.e., 2 ! ways.
$\therefore$ a total number of arrangements when $Y$ and T are always together $=6 ! \times 2$ !
$=720 \times 2$
$=1440$
$\therefore 1440$ words can be formed if $Y$ and $\mathrm{T}$ are together.
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Question 312 Marks
In a group photograph, 6 teachers and principals are in the first row and 18 students are in the second row. There are 12 boys and 6 girls among the students. If the middle position is reserved for the principal and if no two girls are together, find the number of arrangements.
Answer
In 1 st row middle seat is fixed for the principal.
Also 1 st row, 6 teachers can be arranged among themselves in ${ }^6 P_6$ i.e., 6 ! ways. In the 2 nd row, 12 boys can be arranged among themselves in ${ }^{12} \mathrm{P}_{12}$ i.e., 12 ! ways. 13 gaps are created by 12 boys, in which 6 girls are to be arranged. together which can be done in ${ }^{13} \mathrm{P}_6$ ways.
$
\begin{aligned}
& \therefore \text { total number of arrangements }=6 ! \times 12 ! \times{ }^{13} \mathrm{P}_6 \ldots . . .[\text { [using Multiplications Principle] } \\
& =6 ! \times 12 ! \times \frac{13 !}{(13-6) !} \\
& =6 ! \times 12 ! \times \frac{13 !}{7 !} \\
& =\frac{6 ! \times 12 ! \times 13 !}{7 \times 6 !} \\
& =\frac{12 ! 13 !}{7}
\end{aligned}
$
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Question 322 Marks
Determine the number of arrangements of letters of the word ALGORITHM if: $\mathrm{O}$ is first and $\mathrm{T}$ is last.
Answer
A word is to be formed using the letters of the word ALGORITHM.
There are 9 letters in the word ALGORITHM.: When beginning with $\mathrm{O}$ and ends with $\mathrm{T}$ :
All the letters of the word ALGORITHM are to be arranged among themselves such that arrangement begins with $\mathrm{O}$ and ends with $\mathrm{T}$.
7 letters other than $O$ and $T$ can be filled between $O$ and $T$ in ${ }^7 P_7$ i.e., 7 ! ways $=5040$ ways.
$\therefore 5040$ words beginning with $\mathrm{O}$ and ending with $\mathrm{T}$ can be formed.
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Question 332 Marks
Determine the number of arrangements of letters of the word ALGORITHM if:Consonants are at even positions
Answer
A word is to be formed using the letters of the word ALGORITHM.
There are 9 letters in the word ALGORITHM.: When consonants are at even positions:
There are 4 even places and 6 consonants in the word ALGORITHM. 1st, 2nd, 3rd, 4th even places are filled in 6, 5, 4, 3 way respectively.
$\therefore$ The number of ways to fill four even places by consonants $=6 \times 5 \times 4 \times 3=360$ The remaining 5 letters ( 3 vowels and 2 consonants) can be arranged among themselves in ${ }^5 P_5$ i.e., 5 ! ways.
$\therefore$ Total number of ways the words can be formed
In which even places are occupied by consonants $=360 \times 5$ !
$
\begin{aligned}
& =360 \times 120 \\
& =43200
\end{aligned}
$
$\therefore 43200$ words can be formed if even positions are occupied by consonants.
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Question 342 Marks
Determine the number of arrangements of letters of the word ALGORITHM if: no two vowels are together.
Answer
A word is to be formed using the letters of the word ALGORITHM.
There are 9 letters in the word ALGORITHM.: When no two vowels are together:
There are 6 consonants in the word ALGORITHM.
They can be arranged among themselves in ${ }^6 \mathrm{P}_6$ i.e., 6 ! ways.
Let consonants be denoted by $\mathrm{C}$.
${ }_{-} C_{-} C_{-} C_{-} C_{-} C_{-} C_{-}$
6 consonants create 7 gaps in which 3 vowels are to arranged.
$\therefore 3$ vowels can be filled in ${ }^7 \mathrm{P}_3$
$=\frac{7 !}{(7-3) !}$
$=\frac{7 \times 6 \times 5 \times 4 !}{4 !}$
$=210$ ways
$\therefore$ total number of ways in which the word can be formed $=6 ! \times 210$
$=720 \times 210$
$=151200$
$\therefore 151200$ words can be formed if no two vowels are together.
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Question 352 Marks
Determine the number of arrangements of letters of the word ALGORITHM if: vowels are always together.
Answer
A word is to be formed using the letters of the word ALGORITHM.
There are 9 letters in the word ALGORITHM.: When vowels are always together:
There are 3 vowels in the word ALGORITHM. (i.e, A, I, O)
Let us consider these 3 vowels as one unit.
This unit with 6 other letters is to be arranged.
$\therefore$ It becomes an arrangement of 7 things which can be done in ${ }^7 P_7$ i.e., 7 ! ways and 3 vowels can be arranged among themselves in ${ }^3 \mathrm{P}_3$ i.e., 3 ! ways.
$\therefore$ the total number of ways in which the word can be formed $=7 ! \times 3$ !
$=5040 \times 6$
$=30240$
$\therefore 30240$ words can be formed if vowels are always together.
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Question 362 Marks
Find $r$ if ${ }^{12} \mathrm{P}_{\mathrm{r}-2}:{ }^{11} \mathrm{P}_{\mathrm{r}-1}=3: 14$
Answer
$
\begin{aligned}
& { }^{12} \mathrm{P}_{\mathrm{r}-2}:{ }^{11} \mathrm{P}_{\mathrm{r}-1}=3: 14 \\
& \therefore \quad \frac{12 !}{(12-r+2) !} \div \frac{11 !}{(11-r+1) !}=\frac{3}{14} \\
& \therefore \quad \frac{12 !}{(14-r) !} \times \frac{(12-r) !}{11 !}=\frac{3}{14} \\
& \therefore \quad \frac{12 \times 11 !}{(14-r)(13-r)(12-r) !} \times \frac{(12-r) !}{11 !}=\frac{3}{14} \\
& \therefore \quad \frac{12}{(14-r)(13-r)}=\frac{3}{14} \text { MaharashtraBoardSolutions.in } \\
& \therefore(14-r)(13-r)=8 \times 7 \\
&
\end{aligned}
$
Comparing on both sides, we get
$
\begin{aligned}
& 14-r=8 \\
& \therefore r=6
\end{aligned}
$
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Question 372 Marks
Find $n$ if ${ }^n P_6:{ }^n P_3=120: 1$
Answer
Find $n$ if ${ }^n P_6:{ }^n P_3=120: 1$$
\begin{aligned}
& { }^n \mathrm{P}_6:{ }^n \mathrm{P}_3=120: 1 \\
& \therefore \frac{n !}{(n-6) !} \div \frac{n !}{(n-3) !}=\frac{120}{1} \\
& \therefore \frac{\mathrm{n} !}{(\mathrm{n}-6) !} \times \frac{(\mathrm{n}-3) !}{\mathrm{n!}}=120 \\
& \therefore \frac{n !}{(n-6) !} \times \frac{(n-3)(n-4)(n-5)(n-6) !}{n !}=120 \\
& \therefore(\mathrm{n}-3)(\mathrm{n}-4)(\mathrm{n}-5)=120 \\
& \therefore(\mathrm{n}-3)(\mathrm{n}-4)(\mathrm{n}-5)=6 \times 5 \times 4
\end{aligned}
$
Comparing on both sides, we get
$
\begin{aligned}
& \mathrm{n}-3=6 \\
& \therefore \mathrm{n}=9
\end{aligned}
$
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Question 382 Marks
Find n, if: $\frac{n !}{3 !(n-5) !}: \frac{n !}{5 !(n-7) !}=10: 3$
Answer
$
\begin{aligned}
& \frac{\mathrm{n} !}{3 !(n-5) !}: \frac{\mathrm{n} !}{5 !(\mathrm{n}-7) !}=10: 3 \\
& \therefore \quad \frac{\mathrm{n} !}{3 !(\mathrm{n}-5) !} \times \frac{5 !(\mathrm{n}-7) !}{\mathrm{n} !}=\frac{10}{3} \\
& \therefore \quad \frac{\mathrm{n} !}{3 !(\mathrm{n}-5)(\mathrm{n}-6)(\mathrm{n}-7) !} \times \frac{5 \times 4 \times 3 !(\mathrm{n}-7) !}{\mathrm{n} !}=\frac{10}{3} \\
& \therefore \quad \frac{5 \times 4}{(\mathrm{n}-5)(\mathrm{n}-6)}=\frac{10}{3} \text { MaharashtraBoardSolutions.in } \\
& \therefore(\mathrm{n}-5)(\mathrm{n}-6)=3 \times 2
\end{aligned}
$
Comparing on both sides, we get
$
\begin{aligned}
& n-5=3 \\
& \therefore n=8
\end{aligned}
$
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Question 392 Marks
Find $n$, if: $\frac{n !}{3 !(n-3) !}: \frac{n !}{5 !(n-5) !}=5: 3$
Answer
$
\begin{aligned}
& \frac{\mathrm{n} !}{3 !(\mathrm{n}-3) !}: \frac{\mathrm{n} !}{5 !(\mathrm{n}-5) !}=5: 3 \\
& \therefore \quad \frac{\mathrm{n} !}{3 !(\mathrm{n}-3) !} \times \frac{5 !(\mathrm{n}-5) !}{\mathrm{n} !}=\frac{5}{3} \\
& \therefore \quad \frac{\mathrm{n} !}{3 !(\mathrm{n}-3)(\mathrm{n}-4)(\mathrm{n}-5) !} \times \frac{5 \times 4 \times 3 !(\mathrm{n}-5) !}{\mathrm{n} !}=\frac{5}{3} \\
& \therefore \quad \frac{5 \times 4}{(\mathrm{n}-3)(\mathrm{n}-4)}=\frac{5}{3} \text { MaharashtraBoardSolutions.in } \\
& \therefore 12=(\mathrm{n}-3)(\mathrm{n}-4) \\
& \therefore(\mathrm{n}-3)(\mathrm{n}-4)=4 \times 3 \\
&
\end{aligned}
$
Comparing on both sides, we get
$
\begin{aligned}
& n-3=4 \\
& \therefore n=7
\end{aligned}
$
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Question 402 Marks
Show that
$
\frac{(2 n) !}{n !}=2^n(2 n-1)(2 n-3) \ldots 5.3 .1
$
Answer
$
\begin{aligned}
& \text { L.H.S. }=\frac{(2 n) !}{n !} \quad \\
& =\frac{(2 n)(2 n-1)(2 n-2)(2 n-3)(2 n-4) \ldots 6 \times 5 \times 4 \times 3 \times 2 \times 1}{n !} \\
& (2 n)(2 n-1)[2(n-1)](2 n-3)[2(n-2)] \ldots(2 \times 3) \times 5 \\
& =\frac{\times(2 \times 2) \times 3 \times(2 \times 1) \times 1}{n !} \\
& =\frac{2^n[n(n-1)(n-2) \ldots 3.2 .1][(2 n-1)(2 n-3) \ldots 5.3 .1]}{n !} \\
& =\frac{2^{\mathrm{n}}(\mathrm{n} !)(2 \mathrm{n}-1)(2 \mathrm{n}-3) \ldots 5.3 .1}{\mathrm{n} ! } \\
& =2^n(2 n-1)(2 n-3) \ldots .5 .3 .1=\text { R.H.S. } \\
&
\end{aligned}
$
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Question 412 Marks
Show that $\frac{n !}{r !(n-r) !}+\frac{n !}{(r-1) !(n-r+1) !}=\frac{(n+1) !}{r !(n-r+1}$
Answer
$
\begin{aligned}
\text { L.H.S. } & =\frac{n !}{r !(n-r) !}+\frac{n !}{(r-1) !(n-r+1) !} \\
& =\frac{n !}{r(r-1) !(n-r) !}+\frac{n !}{(r-1) !(n-r+1)(n-r) !} \\
& =\frac{n !}{(r-1) !(n-r) !}\left[\frac{1}{r}+\frac{1}{n-r+1}\right] \\
& =\frac{n !}{(r-1) !(n-r) !}\left[\frac{n-r+1+r}{r(n-r+1)}\right] \\
= & \frac{(n+1) \cdot n !}{r \cdot(r-1) !(n-r+1)(n-r) !} \\
= & \frac{(n+1) !}{r !(n-r+1) !}=\text { R.H.S }
\end{aligned}
$
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Question 422 Marks
How many numbers between 100 and 1000 have 4 in the unit’s place?
Answer
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 where the unit place digit is 4.
Since Unit’s place digit is 4.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 10 ways.
For 3-digit number 100’s place digit should be a non-zero number.
∴ 100’s place digit can be selected in 9 ways.
∴ By using fundamental principle of multiplication,
total number of numbers between 100 and 1000 which have 4 in the units place = 1 × 10 × 9 = 90
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Question 432 Marks
In a test that has 5 true/false questions, no student has got all correct answers and no sequence of answers is repeated. What is the maximum number of students for this to be possible?
Answer
For a set of 5 true/false questions, each question can be answered in 2 ways.
∴ By using fundamental principle of multiplication,
the total number of possible sequences of answers = 2 × 2 × 2 × 2 × 2 = 32
Since no student has written all the correct answers.
∴ Total number of sequences of answers given by the students in the class = 32 – 1 = 31
Also, no student has given the same sequence of answers.
∴ Maximum number of students in the class = Number of sequences of answers given by the students = 31
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Question 442 Marks
A letter lock has 3 rings and each ring has 5 letters. Determine the maximum number of trials that may be required to open the lock.
Answer
A letter lock has 3 rings, each ring containing 5 different letters.
∴ A letter from each ring can be selected in 5 ways.
∴ By using fundamental principle of multiplication,
the total number of trials that can be made = 5 × 5 × 5 = 125
Out of these 124 wrong attempts are made and in the 125th attempt,
the lock gets opened, for a maximum number of trials.
∴ A maximum number of trials required to open the lock is 125.
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Question 462 Marks
How many two-letter words can be formed using letters from the word SPACE when repetition of letters
(i) is allowed
(ii) is not allowed
Answer
A two-letter word is to be formed out of the letters of the word SPACE.
(i) When repetition of the letters is allowed
1st letter can be selected in 5 ways
2nd letter can be selected in 5 ways
∴ By using the fundamental principle of multiplication,
total number of 2-letter words = 5 × 5 = 25

(ii) When repetition of the letters is not allowed
1st letter can be selected in 5 ways
2nd letter can be selected in 4 ways
∴ By using the fundamental principle of multiplication,
total number of 2-letter words = 5 × 4 = 20

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Question 472 Marks
A Signal is generated from 2 flags by putting one flag above the other. If 4 flags of different colours are available, how many different signals can be generated?
Answer
A signal is generated from 2 flags and there are 4 flags of different colours available.
∴ 1st flag can be any one of the available 4 flags.
∴ It can be selected in 4 ways.
Now, 2nd flag is to be selected for which 3 flags are available for a different signal.
∴ 2nd flag can be anyone from these 3 flags.
∴ It can be selected in 3 ways.
∴ By using the fundamental principle of multiplication,
Total number of ways in which a signal can be generated = 4 × 3 = 12
∴ 12 different signals can be generated.
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Question 482 Marks
In question 1, in how many ways can the monitor be selected if the monitor must be a boy? What is the answer if the monitor must be a girl?
Answer
(i) Since there are 30 boys in the class
∴ A boy monitor can be selected in 30 ways.
(ii) Since there are 20 girls in the class
∴ A girl monitor can be selected in 20 ways.
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Question 492 Marks
How many four-digit numbers will not exceed 7432 if they are formed using the digits 2, 3, 4, 7 without repetition?
Answer
Among many set’s of digits, the greatest number is possible when digits are arranged in descending order.
∴ 7432 is the greatest number, formed from the digits 2, 3, 4, 7.
∴ Since a 4-digit number is to be formed from the digits 2, 3, 4, 7, where repetition of the digit is not allowed.
∴ 1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
∴ Total number of numbers not exceeding 7432 that can be formed from the digits 2, 3, 4, 7
= Total number of four-digit numbers formed from the digits 2, 3, 4, 7
= 4 × 3 × 2 × 1
= 24
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