MCQ 1511 Mark
Eight straight lines are drawn in the plane such that no two lines are parallel and no three lines are concurrent.The number of parts into which these lines divide the plane, is:
MCQ 1521 Mark
If $C_0+ C_1 + C_2 + ... + C_n = 256$, then $^{2n}C_2$ is equal to:
AnswerIf set $S$ has $n$ elements, then $\text{C (n, k)C n, k}$ is the number of ways of choosing $k$ elements from $S.$
Thus, the number of subsets of $\text{SS}$ of all possible values is given by,
$\text{C}(\text{n},0)+\text{C}(\text{n},1)+\text{C}(\text{n},3)+.....+\text{C}(\text{n},\text{n})=2^\text{n}$
Comparing the given equation with the above equation:
$2^\text{n}=256$
$\Rightarrow 2^\text{n}=2^{8}$
$\Rightarrow \text{n}=8$
$\therefore {^\text{2n}}\text{C}_{\text{2}}={^\text{16}}\text{C}_{\text{2}}$
$\Rightarrow {^\text{16}}\text{C}_{\text{2}}=\frac{16!}{2!4!}=\frac{16\times15}{2}=120$
View full question & answer→MCQ 1531 Mark
Choose the correct answer. The total number of $9$ digit numbers which have all different digits is.
- A
$10!$
- B
$9 !$
- ✓
$9 \times 9!$
- D
$10\times 10!$
AnswerCorrect option: C. $9 \times 9!$
We have to form $9-$digit number which has all different digit.
First digit from the left can be filled in $9$ ways $($excluding $'0\ ').$
Now nine digits are left including $'O\ '.$
So remaining eight places can be filled with these nine digits in ${ }^9 P_S$ ways.
So, total number of numbers $=9 \times{ }^9 \mathrm{P}_8=9 \times 9$ !
View full question & answer→MCQ 1541 Mark
The number of ways in which a host lady can invite for a party of $8$ out of $12$ people of whom two do not want to attend the party together is:
- A
$2\times{^\text{11}}\text{C}_{\text{7}}+{^\text{10}}\text{C}_{\text{8}}$
- B
${^\text{10}}\text{C}_{\text{8}}+{^\text{11}}\text{C}_{\text{7}}$
- ✓
${^\text{12}}\text{C}_{\text{8}}-{^\text{10}}\text{C}_{\text{6}}$
- D
AnswerCorrect option: C. ${^\text{12}}\text{C}_{\text{8}}-{^\text{10}}\text{C}_{\text{6}}$
A host lady can invite $8$ out of $12$ people in ways.
Two out of these $12$ people do not want to attend the party together.
Number of ways $={^\text{12}}\text{C}_{\text{8}}-{^\text{10}}\text{C}_{\text{6}}.$
View full question & answer→MCQ 1551 Mark
In how many ways can we paint the six faces of a cube with six different colours:
AnswerNow, let us consider the $6$ different colours: $c^1, c^2, c^3, c^4, c^5, c^6$
Assume that the face of the cube facing up is $c^1$
So, the face of the cube at the bottom can be painted in $5$ different ways.
So, $4$ faces on the horizontal side of the cube are in circular permutation and they can be painted in $(4 - 1)!$ ways.
Hence, the total number of ways we can paint the faces of a cube with six different colours is $5 \times (4 - 1)!$ ways.
Hence, $5 \times (4 - 1)!$ ways $= 5 \times 3! = 5 \times 3 \times 2 \times 1 = 30$ ways.
View full question & answer→MCQ 1561 Mark
If ${^\text{20}}\text{C}_{3\text{r+1}}={^\text{20}}\text{C}_{\text{r-1}},$ is then $r$ equal to:
Answer$\text{r}+\text{1}+\text{r}-1=20$ $[\therefore \ ^\text{n}\text{C}_{\text{x}} = \ ^\text{n}\text{C}_{\text{y}} \Rightarrow \text{n} = \text{x} + \text{y} \text{ or } \text{x = y}]$
$\Rightarrow 2\text{r}=20$
$\Rightarrow \text{r}=10$
View full question & answer→MCQ 1571 Mark
The value of $ 2^\text{n}\big[1·3· 5…(2\text{n}-3)2\text{n}-1\big)$ is:
- ✓
$\frac{(2\text{n} )!}{\text{n!}}$
- B
$\frac{(2\text{n})!}{2^\text{n}}$
- C
$\frac{\text{n}!}{( 2\text{n})!}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\frac{(2\text{n} )!}{\text{n!}}$
View full question & answer→MCQ 1581 Mark
Choose the correct answer.
The sum of the digits in unit place of all the numbers formed with the help of $3, 4, 5$ and $6$ taken all at a time is.
AnswerIf the unit place is $'3\ '$ then remaining three places can be filled in $3!$ ways.
Thus $'3\ '$ appears in unit place in $3!$ times.
Similarly each digit appear in unit place $3!$ times.
So, sum of digits in unit place $= 3! (3 + 4 + 5 + 6) = 18 \times 6 = 108$
View full question & answer→MCQ 1591 Mark
The products of any $r$ consecutive natural numbers is always divisible by:
View full question & answer→MCQ 1601 Mark
In how many ways can the letters of the word $\text{CORPORATION}$ be arranged so that vowels always occupy even places:
- A
$120$
- B
$2700$
- C
$720$
- ✓
$7200$
AnswerCorrect option: D. $7200$
View full question & answer→MCQ 1611 Mark
A garrison of $'n\ ' $men had enough food to last for $30$ days. After $10$ days, $50$ more men joined them. If the food now lasted for $16$ days, what is the value of $n?$
AnswerAfter $10$ days, the food for $n$ men is there for $20$ days.
This food can be eaten by $(n + 50)$ men in $16$ days.
$\therefore 20n = 16(n + 50)$
$\therefore n = 200$
View full question & answer→MCQ 1621 Mark
If $^\text{ⁿ}\text{C}_{15} = \ ^\text{ⁿ}\text{C}_6$ then the value of $\ ^\text{ⁿ}\text{C}_{21}$ is:
AnswerWe know that
if $\ ^\text{ⁿ}\text{Cr}_1 =\ ^ⁿ\text{Cr}_2$
$\Rightarrow\text{n}= \text{r}^1 + \text{r}^2$
Given, $^\text{ⁿ}\text{C}_{15} = \ ^\text{ⁿ}\text{C}_6 $
$\Rightarrow\text{n} = 15 + 6$
$\Rightarrow\text{n} = 21$
Now, $ ^{21}\text{C}_{21} = 1$
View full question & answer→MCQ 1631 Mark
$3$ letters are posted in $5$ letters boxes. If all the letters are not posted in the same box, then number of ways of posting is:
AnswerAccording to problem, $3$ letters are posted $5$ boxesimplies that, we have a letter and can be posted in any of the $5$ boxes.
Similarly next letter can be posted in $5$ ways, and the all other follow same methodimplies that.
$(5)^3=5\times5\times5=125$
View full question & answer→MCQ 1641 Mark
The formula for permutations and combinations are related as:$ \ ^\text{n}\text{C}_\text{r} =\frac {\ ^\text{n}\text{P}\text{r}}{\text{r}!}$
AnswerThe formula for permutations and combinations are related as:
$\ ^\text{n}\text{C}_\text{r} =\frac {\ ^\text{n}\text{P}\text{r}}{\text{r}!}$
View full question & answer→MCQ 1651 Mark
Choose the correct answer.
Every body in a room shakes hands with everybody else. The total number of hand shakes is $66.$ The total number of persons in the room is.
AnswerBetween any two person there is one hand shake.
So, number of hand shakes $=\ ^\text{n}\text{C}_2=\frac{\text{n!}}{2!(\text{n}-2)!}=\frac{\text{n}(\text{n}-1)}{2}=66($given$)$
$\Rightarrow\text{n}(\text{n}-1)=132$
$\Rightarrow\text{n}^2-\text{n}-132=0$
$\Rightarrow(\text{n}-12)(\text{n}=11)=0$
$\therefore\text{n}=12$
View full question & answer→MCQ 1661 Mark
Among $14$ players, $5$ are bowlers. In how many ways a team of $11$ may be formed with at least $4$ bowlers?
AnswerAmong $14$ players, $5$ are bowlers.
A team of $11$ players has to be selected such that at least $4$ bowlers are included in the team.
Required number of ways $={^\text{5}}\text{C}_{\text{4}}\times{^\text{9}}\text{C}_{\text{7}}+{^\text{5}}\text{C}_{\text{5}}\times{^\text{9}}\text{C}_{\text{6}}$
$=180+84$
$=264$
View full question & answer→MCQ 1671 Mark
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is:
AnswerA parallelogram can be formed by choosing two parallel lines from the set of four parallel lines and two parallel lines from the set of three parallel lines.
Two parallel lines from the set of four parallel lines can be chosen in ${^\text{4}}\text{C}_{\text{2}}$ ways and two parallel lines from the set of $3$ parallel lines can be chosen in ${^\text{3}}\text{C}_{\text{2}}$ ways.
Number of parallelograms that can be formed
$={^\text{4}}\text{C}_{\text{2}}\times{^\text{3}}\text{C}_{\text{2}}$
$=\frac{4!}{2!2!}\times\frac{3!}{2!1!}$
$=6\times3$
$=18$
View full question & answer→MCQ 1681 Mark
If a represents the number of permutations of $(x + 2)$ things taken together b represents the number of permutation of $11$ things taken together out of $x$ things, and $c$ represents the number of permutation of $(x – 11)$ things taken together so that $a = 182, bc =$ then $x$ is equal to:
View full question & answer→MCQ 1691 Mark
Choose the correct answer. The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is.
AnswerTo form parallelogram we required a pair of line from a set of $4$ lines and another pair of line from another set of $3$ lines.
Required number of parallelograms $= { }^4 C_2 \times{ }^3 C_2=6 \times 3=18$
View full question & answer→MCQ 1701 Mark
A committee of $7$ has to be formed from $9$ boys and $4$ girls. In how many ways can this be done when the committee consists of exactly $3$ girls:
AnswerGiven number of boys $= 9$
Number of girls $= 4$
Now, A committee of $7$ has to be formed from $9$ boys and $4$ girls.
Now, If in committee consist of exactly $3$ girls:
$\ ^4\text{C}_3\times\ ^9\text{C}_4$
$= \bigg(\frac{4!}{(3! \times 1!)}\bigg)\times\bigg(\frac{9!}{(4!\times5!)}\bigg)$
$= \bigg(\frac{4\times3}{(3!)}\bigg)\times\bigg(\frac{9\times8\times7\times6\times5!}{(4!\times5!)}\bigg)$
$= 4\times\frac{(9\times 8\times7\times6)}{4!}$
$= 4\times\frac{(9\times 8\times7\times6)}{(4\times3\times2\times1)}$
$=9\times8\times7$
$= 504$
View full question & answer→MCQ 1711 Mark
On the occasion of Deepawali festival, each student of a class sends greeting cards to the others.If there are $20$ students in the class, then the total number of greeting cards exchanged by the students is:
- A
$\ ^{20}\text{C}_2$
- ✓
$2\ ^{20}\text{C}_2$
- C
$2\ ^{20}\text{P}_2$
- D
$\text{None of these}$
AnswerCorrect option: B. $2\ ^{20}\text{C}_2$
View full question & answer→MCQ 1721 Mark
In how many ways $8$ distinct toys can be distributed among $5$ children?
- A
$\ ^8\text{P}_5$
- B
$\ ^5\text{P}_8$
- ✓
$5^8$
- D
$8^5$
AnswerGiven that, the number of toys $ = 8$
The number of children $ = 5.$
Hence, the number of ways $8$ distinct toys can be distributed among $5$ children is $5\times5\times5\times5\times5\times5\times5\times5 =5^8.$
View full question & answer→MCQ 1731 Mark
Total number of words formed by $2$ vowels and $3$ consonants taken from $4$ vowels and $5$ consonants is equal to:
AnswerCorrect option: C. $7200$
$2$ out of $4$ vowels can be chosen in ${ }^4 C_2$ ways and $3$ out of $5$ consonants can be chosen in ${ }^5 C_3$ ways.
Thus, there are $({^\text{4}}\text{C}_{\text{2}}\times{^\text{5}}\text{C}_{\text{3}})$ groups, each containing $2$ vowels and $3$ consonants.
Each group contains $5$ letters that can be arranged in $5!$ ways.
Required number of words $=({^\text{4}}\text{C}_{\text{2}}\times{^\text{5}}\text{C}_{\text{3}})\times5!$
$=60\times120$
$=7200$
View full question & answer→MCQ 1741 Mark
Given $11$ points, of which $5$ lie on one circle, other than these $5$, no $4$ lie on one circle. Then the number of circles that can be drawn so that each contains at least $3$ of the given points is:
AnswerWe need at least three points to draw a circle that passes through them.
Now, number of circles formed out of $11$ points by taking three points at a time $={^\text{11}}\text{C}_{\text{3}}=165$
Number of circles formed out of $5$ points by taking three points at a time $={^\text{5}}\text{C}_{\text{3}}=10$
It is given that $5$ points lie on one circle.
Required number of circles $= 165 - 10 + 1 = 156.$
View full question & answer→MCQ 1751 Mark
A person wishes to make up as many different parties as he can out of $20$ friends. Each party consists of the same number of friends. How many should be invited at a time:
View full question & answer→MCQ 1761 Mark
Number of all four digit numbers having different digits formed of the digits $1, 2, 3, 4$ and $5$ and divisible by $4$ is:
AnswerIn order to make a number divisible by $4$, its last two digits must be divisible by $4,$ which in this case can be $12, 24, 32$ or $52.$
Since repetition of digits is not allowed, the remaining first two digits can be arranged in $3 \times 2$ ways in each case.
$\therefore$ Total number of numbers that can be formed $= 4 \times {3 \times 2} = 24$
View full question & answer→MCQ 1771 Mark
If $C(n, 12) = C(n, 8)$ is then $r$ equal to:
Answer${^\text{n}}\text{C}_{\text{12}}={^\text{n}}\text{C}_{\text{8}}$
$\Rightarrow \text{n}=12+8=20$
${^\text{n}}\text{C}_{\text{x}}={^\text{n}}\text{C}_{\text{y}}$
$\Rightarrow \text{n}=\text{x}+\text{y}, \text{x}=\text{y}$
Now,
${^\text{22}}\text{C}_{\text{n}}={^\text{22}}\text{C}_{\text{20}}$
$=\frac{22}{2}\times\frac{21}{1}$
$=231$
View full question & answer→MCQ 1781 Mark
Value of $0!$ is always $1:$
Answerwe know $1! = 1$
Also $n! = n \times (n − 1) \times (n − 2)........3 \times 2 \times 1n!$
$= n \times (n − 1)! 1 !$
$= 1(1 − 1)! 1$
$= 1(0)! 0 !$
$= 1$
$0!$ is always $1$
View full question & answer→MCQ 1791 Mark
Let $R = (a, b, c, d)$ and $S = (1, 2, 3),$ then the number of functions $f,$ from $R$ to $S,$ which are onto is:
AnswerTotal number of functions $= 3^4 = 81$
All the four elements can be mapped to exactly one element in $3$ ways, and exactly $3$
elements in $3(2^4- 2) = 3(16 - 2) = 3 \times 14 = 42$
Thus the number of onto functions $= 81 - 42 - 3 = 81 - 45 = 36$
View full question & answer→MCQ 1801 Mark
The number of six letter words that can be formed using the letters of the word $"\text{ASSIST}"$ in which $\text{S's}$ alternate with other letters is:
AnswerAll $S's$ can be placed either at even places or at odd places,
i.e. in $2$ ways. The remaining letters can be placed at the remaining places in $3!,$
i.e. in $6$ ways.
$\therefore$ Total number of ways $= 6 \times 2 \times 2 = 12$
View full question & answer→MCQ 1811 Mark
In how many ways can $10$ lion and $6$ tigers be arranged in a row so that no two tigers are together?
- ✓
$10!\times\ ^{11}\text{p}_6$
- B
$10!\times\ ^{10}\text{p}_6$
- C
$6!\times\ ^{10}\text{p}_7$
- D
$6!\times\ ^{10}\text{p}_6$
AnswerCorrect option: A. $10!\times\ ^{11}\text{p}_6$
View full question & answer→MCQ 1821 Mark
A letter lock contains $5$ rings each marked with for different letters.The number of all possible unsuccessful attempts to open the lock is:
- A
$625$
- ✓
$1024$
- C
$624$
- D
$1023$
AnswerCorrect option: B. $1024$
View full question & answer→MCQ 1831 Mark
The number of words from the letters of the word $'\text{BHARAT}'$ in which $B$ and $H$ will never come together, is:
AnswerTotal number of words that can be formed of the letters of the word $\text{BHARAT} =\frac{6!}{2!}$
Number of words in which the letters $B$ and $H$ are always together $=2\times\frac{5!}{2!}$
$=120$
$\therefore$ Number of words in which the letters $B$ and $H$ are never together $=360-120$
$=240$
View full question & answer→MCQ 1841 Mark
If ${^\text{n}}\text{C}_{\text{12}}={^\text{n}}\text{C}_{\text{8}},$ is then $n:$
View full question & answer→MCQ 1851 Mark
In a class there are $18$ boys who are over $160 \ cm$ tall If these constitute three$-$fourths of the boys and the total number of boys is tow$-$third of the total number of students in the class what is the number of girls in the class?
AnswerGiven in class there are $18$ boys who over $160 \ cm$ tall and these are three- fourths of total boys
Then total boys in class $=18\div\frac{3}{4}=18\times\frac{4}{3}=24$
And total no of student in class $=24\times\frac{3}{2}=36$
Then number of girls $=36-24=12$
View full question & answer→MCQ 1861 Mark
The value of $^6C_4$ is:
Answer$^6\text{c}_4=^6\text{c}_{6-4}=^6\text{c}_2=\frac{6\times5}{2\times1}=15$
View full question & answer→