MCQ 11 Mark
If ${^\text{20}}\text{C}_{\text{r}}={^\text{20}}\text{C}_{\text{r+4}}$ is then ${^\text{r}}\text{C}_{\text{3}}$ equal to :
Answer$\text{r}+\text{r}+4=20$
$\Rightarrow 2\text{r}+4=20$
$\Rightarrow 2\text{r}=16$
$\Rightarrow \text{r}=8$
Now,
${^\text{r}}\text{C}_{\text{3}}={^\text{8}}\text{C}_{\text{3}}$
$\therefore\ {^\text{8}}\text{C}_{\text{3}}={^\text{8}}\text{C}_{\text{3}}$
$\therefore\ {^\text{8}}\text{C}_{\text{3}}=\frac{8!}{3!5!}$
$=\frac{8\times7\times6}{3\times2\times1}=56$
View full question & answer→MCQ 21 Mark
The value of $({^\text{7}}\text{C}_{\text{0}}+{^\text{7}}\text{C}_{\text{1}})+({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{3}})+.....+({^\text{7}}\text{C}_{\text{6}}+{^\text{7}}\text{C}_{\text{7}})$ is :
- A
$2^{7}-1$
- ✓
$2^{8}-2$
- C
$2^{8}-1$
- D
$2^{8}$
AnswerCorrect option: B. $2^{8}-2$
$({^\text{7}}\text{C}_{\text{0}}+{^\text{7}}\text{C}_{\text{1}})+({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{3}})+({^\text{7}}\text{C}_{\text{2}}+{^\text{7}}\text{C}_{\text{3}})+({^\text{7}}\text{C}_{\text{3}}+{^\text{7}}\text{C}_{\text{4}})+......$
$=1+2\times{^\text{7}}\text{C}_{\text{1}}+2\times{^\text{7}}\text{C}_{\text{2}}+2\times{^\text{7}}\text{C}_{\text{3}}+2\times{^\text{7}}\text{C}_{\text{4}}+2\times{^\text{7}}\text{C}_{\text{5}}..$
$=2+2^{2}({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{2}}+{^\text{7}}\text{C}_{\text{3}})$
$=2+2^{2}(7+\frac{7}{2}\times6+\frac{7}{3}\times\frac{6}{2}\times5)$
$=2+252$
$=254$
$=2^{8}-2$
View full question & answer→MCQ 31 Mark
Three persons enter a railway compartment. If there are $5$ seats vacant, in how many ways can they take these seats?
AnswerThree persons can take $5$ seats in $^5C_3$ ways. Moreover $3$ persons can sit in $3!$ ways.
Required number of ways ${^\text{5}}\text{C}_{\text{3}}\times3!$
$=10\times6=60$
View full question & answer→MCQ 41 Mark
If ${^\text{15}}\text{C}_{3\text{r}}={^\text{15}}\text{C}_{\text{r+3}},$ is then equal to :
Answer$3\text{r}+\text{r}+3=15$
$\Rightarrow 4\text{r}+3=15$
$\Rightarrow 4\text{r}=12$
$\Rightarrow \text{r}=3$
View full question & answer→MCQ 51 Mark
If ${^\text{20}}\text{C}_{\text{r}}={^\text{20}}\text{C}_{\text{r-10}}$ is then ${^\text{18}}\text{C}_{\text{r}}$ equal to :
Answer$\text{r}+\text{r}-10=20$
$\Rightarrow 2\text{r}-10=20$
$\Rightarrow 2\text{r}=30$
$\Rightarrow \text{r}=15$
Now,
${^\text{18}}\text{C}_{\text{r}}={^\text{18}}\text{C}_{\text{15}}$
$\therefore\ {^\text{18}}\text{C}_{\text{15}}={^\text{18}}\text{C}_{\text{3}}$
$\therefore\ {^\text{18}}\text{C}_{\text{3}}=\frac{18}{3}\times\frac{17}{2}\times16$
$=816$
View full question & answer→MCQ 61 Mark
If $^{(\text{a}^2-\text{a})}\text{C}_{\text{2}}=^{(\text{a}^2-\text{a})\text{}}\text{C}_{\text{4}},$ is then $x : Z$
Answer$\text{a}^{2}-\text{a}=2+4$
$\Rightarrow \text{a}^{2}-\text{a}-6=0$
$\Rightarrow \text{a}^{2}-3\text{a}+2\text{a}-6=0$
$\Rightarrow \text{a}(\text{a}-3)+2(\text{a}-3)=0$
$\Rightarrow (\text{a}+2)(\text{a}-3)=0$
$\Rightarrow \text{a}=-2,\text{a}=3$
$\text{a}=3$
View full question & answer→MCQ 71 Mark
If $={^\text{43}}\text{C}_{\text{r-6}}={^\text{43}}\text{C}_{\text{3r+1}},$ then the value of $r$ is is :
Answer$\text{r}-6+3\text{r}+1=43$
$\Rightarrow 4\text{r}-5=43$
$\Rightarrow 4\text{r}=48$
$\Rightarrow \text{r}=12$
View full question & answer→MCQ 81 Mark
A lady gives a dinner party for six guests. The number of ways in which they may be selected from among ten friends if two of the friends will not attend the party together is :
AnswerSuppose there are two friends, $A$ and $B,$ who do not attend the party together.
If both of them do not attend the party, then the number of ways of selecting $6$ guests $={^\text{8}}\text{C}_{\text{6}}=28$
If one of them attends the party, then the number of ways of selecting $6$ guests $=2.{^\text{8}}\text{C}_{\text{5}}=112$
Total number of ways $= 112 + 28 = 140$
View full question & answer→MCQ 91 Mark
There are $13$ players of cricket, out of which $4$ are bowlers. In how many ways a team of eleven be selected from them so as to include at least two bowlers?
Answer$4$ out of $13$ players are bowlers.
In other words, $9$ players are not bowlers.
A team of $11$ is to be selected so as to include at least $2$ bowlers.
Number of ways $={^\text{4}}\text{C}_{\text{2}}\times{^\text{9}}\text{C}_{\text{9}}+{^\text{4}}\text{C}_{\text{3}}\times{^\text{9}}\text{C}_{\text{8}}+{^\text{4}}\text{C}_{\text{4}}\times{^\text{9}}\text{C}_{\text{7}}$
$=6+36+36$
$=78$
View full question & answer→MCQ 101 Mark
${^\text{5}}\text{C}_{\text{1}}+{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{3}}+{^\text{5}}\text{C}_{\text{4}}+{^\text{5}}\text{C}_{\text{5}}$is equal to :
Answer${^\text{5}}\text{C}_{\text{1}}+{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{3}}+{^\text{5}}\text{C}_{\text{4}}+{^\text{5}}\text{C}_{\text{5}}$
$={^\text{5}}\text{C}_{\text{1}}+{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{1}}+{^\text{5}}\text{C}_{\text{5}}$
$=2\times{^\text{5}}\text{C}_{\text{1}}+2\times{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{5}}$
$= 2\times5+2\times\frac{5!}{2!3!}+1$
$=10+20+1$
$=31$
View full question & answer→MCQ 111 Mark
How many different committees of $5$ can be formed from $6$ men and $4$ women on which exact $3$ men and $2$ women serve?
AnswerNumber of committes that can be formed $={^\text{6}}\text{C}_{\text{3}}\times{^\text{4}}\text{C}_{\text{2}}$
$=\frac{6!}{3!3!}\times\frac{4!}{2!2!}$
$=\frac{6\times5\times4}{3\times2}\times\frac{4\times3}{2}$
$=120$
View full question & answer→MCQ 121 Mark
In how many ways can a committee of $5$ be made out of $6$ men and $4$ women containing at least one women?
AnswerRequired number of ways $={^\text{4}}\text{C}_{\text{1}}\times{^\text{6}}\text{C}_{\text{4}}+{^\text{4}}\text{C}_{\text{2}}\times{^\text{6}}\text{C}_{\text{3}}+{^\text{4}}\text{C}_{\text{3}}\times{^\text{6}}\text{C}_{\text{2}}+{^\text{4}}\text{C}_{\text{4}}\times{^\text{6}}\text{C}_{\text{1}}$
$=60+120+60+6$
$=246$
View full question & answer→MCQ 131 Mark
There are $10$ points in a plane and $4$ of them are collinear. The number of straight lines joining any two of them is :
AnswerNumber of straight lines formed by joining the $10$ points if we take $2$ points at a time $={^\text{10}}\text{C}_{\text{2}}=\frac{10}{2}\times\frac{9}{1}=45$
Number of straight lines formed by joining the $4$ points if we take $2$ points at a time $={^\text{4}}\text{C}_{\text{2}}=\frac{4}{2}\times\frac{3}{1}=6$
But, $4$ collinear points, when joined in pairs, give only one line.
Required number of straight lines $= 14 - 6 + 1 = 40$
View full question & answer→MCQ 141 Mark
If ${^\text{m}}\text{C}_{\text{1}}={^\text{n}}\text{C}_{\text{2}},$ is then :
- A
$2m = n$
- B
$2m = n(n + 1)$
- ✓
$2m = n(n - 1)$
- D
$2n = m(m - 1)$
AnswerCorrect option: C. $2m = n(n - 1)$
${^\text{m}}\text{C}_{\text{1}}={^\text{n}}\text{C}_{\text{2}}$
$\Rightarrow \frac{\text{m!}}{1!(\text{m}-1)!}$
$=\frac{\text{n!}}{2!(\text{n}-2)!}$
$\Rightarrow \frac{\text{m}(\text{m}-1)!}{(\text{m}-1)!}$
$=\frac{\text{n}(\text{n}-1)(\text{n}-2)!}{2!(\text{n}-2)!}$
$\Rightarrow2\text{m}=\text{n}(\text{n}-1)$
View full question & answer→MCQ 151 Mark
If ${^\text{n+1}}\text{C}_{\text{3}}=2.{^\text{n}}\text{C}_{\text{2}},$ then $n$ :
Answer${^\text{n+1}}\text{C}_{\text{3}}=2\times{^\text{n}}\text{C}_{\text{2}}$
$\Rightarrow \frac{(\text{n}+1)!}{3!(\text{n-2})!}=2\times\frac{\text{n}!}{2!(\text{n}-1)!}$
$\Rightarrow \text{n+1}=6$
$\Rightarrow \text{n}=5$
View full question & answer→MCQ 161 Mark
There are $12$ points in a plane. The number of the straight lines joining any two of them when $3$ of them are collinear is :
AnswerNumber of straight lines joining $12$ points if we take $2$ points at a time $={^\text{12}}\text{C}_{\text{2}}$
$=\frac{12!}{2!10!}=66$
Number of straight lines joining $3$ points if we take $2$ points at a time $={^\text{3}}\text{C}_{\text{2}}=3$
Required number of straight lines $= 66 - 3 + 1 = 64$
View full question & answer→MCQ 171 Mark
If ${^\text{n}}\text{C}_{\text{r}}+{^\text{n}}\text{C}_{\text{r+1}}={^\text{n+1}}\text{C}_{\text{x}},$ is then $x$ :
- A
$\text{r}$
- B
$\text{r}-1$
- C
$\text{n}$
- ✓
$\text{r}+1$
AnswerCorrect option: D. $\text{r}+1$
We have,
${^\text{n}}\text{C}_{\text{r}}+{^\text{n}}\text{C}_{\text{r+1}}={^\text{n+1}}\text{C}_{\text{x}}$
$\Rightarrow {^\text{n+1}}\text{C}_{\text{r+1}}={^\text{n+1}}\text{C}_{\text{x}}$
$\Rightarrow \text{r}+1=\text{x}$
${^\text{n}}\text{C}_{\text{x}}={^\text{n}}\text{C}_{\text{y}}$
$\Rightarrow \text{n}=\text{x}+\text{y},\text{x}=\text{y}$
View full question & answer→MCQ 181 Mark
The number of diagonals that can be drawn by joining the vertices of an octagon is :
AnswerAn octagon has $8$ vertices.
The number of diagonals of a polygon is given by $\frac{\text{n}(\text{n}-3)}{2}$
Number of diagonals of an octagon $=\frac{\text{8}(\text{8}-3)}{2}=20$
View full question & answer→MCQ 191 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 $C (n, k)C n, k$ is the number of ways of choosing $k$ elements from $S.$
Thus, the number of subsets of $SS$ 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 201 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 211 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 221 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 231 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 241 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 $^4C_2$ ways and $3$ out of $5$ consonants can be chosen in $^5C_3$ 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 251 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 261 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 271 Mark
If ${^\text{n}}\text{C}_{\text{12}}={^\text{n}}\text{C}_{\text{8}},$ is then $n$ :
View full question & answer→