MCQ Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 3

$^nC_3 : ^{2n-1}C_2 =$

$8:15$ , then

$n$

$=$

0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

$^nC_3 : ^{2n-1}C_2 = 8:15$

$\Rightarrow \displaystyle 15\cdot ^nC_3= 8\cdot ^{2n-1}C_2$

$\Rightarrow \displaystyle 15\cdot \frac{n(n-1)(n-2)}{3!}=8\cdot \frac{(2n-1)(2n-2)}{2!}$

$\Rightarrow \displaystyle 15\cdot \frac{n(n-1)(n-2)}{6}=8\cdot \frac{2(2n-1)(n-1)}{2}$

$\Rightarrow \displaystyle 5\cdot \frac{n(n-2)}{2}=8\cdot \frac{(2n-1)}{1}$

$\Rightarrow 5n(n-2) = 16(2n-1)$

$\Rightarrow 5n^2-42n+16=0$

$\Rightarrow n=8$
Note: Easiest way will be is to put values of

$'n'$ and check options

The least value of the natural number '

$n$ ' satisfying

$c(n,5) + c(n,6) > c(n+1,5)$

0%
10
0%
12
0%
13
0%
11

Explanation

The least value of

$n$

$^{n}C_{5}+^{n}C_{6}=^{(n+1)}C_{6}> ^{(n+1)}C_{5}$

$\Rightarrow n+1> 6+5$

$\Rightarrow n=11$

Match the following :

$A)^{ n }C_{ 0 }+^{ n }C_{ 1 }+^{ n }C_{ 2 }+....+^{ n }C_{ n }$	1) $^{ (n+1) }P_{ r }$
$B) { \dfrac { P_{ r } }{ P_{ (r-1) } } } \;\;\; \;$	2) $2^n$
$C) ^nP_r + r ^n .P_{(r-1)}$	3) $^{(n+1)}P_{r}$
$D) ^nP_n$	4) ${n-r+1}$
	5) $n!$

0%
$A-2, B-4,C-1,D-5$
0%
$A-1, B-4,C-2,D-5$
0%
$A-2, B-4,C-5,D-1$
0%
$A-1, B-3,C-5,D-2$

Explanation

$^{n}C_{0}+^{n}C_{1}+.....^{n}C_{n}=(1+1)^n=2^{n}$

$\cfrac{^{n}P_{r}}{^{n}P_{r-1}}=\cfrac{n!(n-r+1)!}{n!(n-r)!}=n-r+1$

$^nP_r+r. ^nP_{r-1}=\cfrac{n!}{(n-r)!}+r.\cfrac{n!}{(n-r+1)!}=\cfrac{n!}{(n-r+1)!}(n-r+1+r)=\cfrac{(n+1)!}{(n-r+1)!}=^{n+1}P_r$

$^nP_n = \cfrac{n!}{(n-n)!}=n!$

$^nC_r + ^nC_{r+1} = ^{(n+1)}C_x$ , then

$x=$

0%
$r$
0%
$r-1$
0%
$n$
0%
$r+1$

Explanation

$\displaystyle ^{n}C_{r}+^{n}C_{r+1}=\frac{n!}{(n-r)!r!}+\frac{n!}{(n-r-1)!(r+1)!}$

$\quad \displaystyle =\frac{n!}{(n-r)!(r+1)!}\left(r+1+n-r \right)=\frac{n!}{(n-r)!(r+1)!}\left(n+1 \right)$

$\quad \displaystyle =\frac{(n+1)!}{(n-r)!(r+1)!}= ^{(n+1)}C_{r+1}$

$\therefore x=r+1$

$^nP_r = 840, ^nC_r = 35$ , then

$n$ is equal to:

0%
6
0%
7
0%
8
0%
9

Explanation

$\dfrac{^{n}P_{r}}{^{n}C_{r}}=r!=\dfrac{840}{35}=24\Rightarrow r=4$

$^{n}P_{r}= \dfrac{n!}{(n-r)!}=840$

$\Rightarrow n \times (n-1) \times (n-2) \times (n-3)=840$

$\Rightarrow 840= 4\times 5 \times 6 \times 7$

$\Rightarrow n=7$

$C (2n, 3) : C (n, 2)$

$=$

$12 : 1$ , then

$n$

$=$

0%
$4$
0%
$5$
0%
$6$
0%
$8$

Explanation

We have,

$C(2n,3):C(n,2)=12:1$

$\Rightarrow \dfrac {^{2n}C_3}{^nC_2}=\dfrac {12}{1}$

$\Rightarrow ^{2n}C_3=12\cdot \ ^nC_2$

$\Rightarrow \displaystyle \frac{2n(2n-1)(2n-2)}{3!}=12\cdot \frac{n(n-1)}{2!}$

$\Rightarrow \displaystyle \frac{4(2n-1)}{6}=\frac{12}{2}$

$\Rightarrow 2n-1=9$

$\Rightarrow n=5$

$^{n+2}C_8 : ^{n-1}P_4 = 133:48$ , then

$n$

$=$

0%
$17$
0%
$18$
0%
$19$
0%
$20$

Explanation

$^{n+2}C_8 : ^{n-1}P_4 = 133:48$

$\displaystyle \Rightarrow \frac{(n+2)!}{(n-6)!8!}.\frac{(n-5)!}{(n-1)!}=\frac{133}{48}$

$\Rightarrow (n-5)(n)(n+1)(n+2)=19\times 7\times 6\times 5\times 4 \times 7=14.19.20.21$

$\therefore n = 19$

$\left(\:^{22}C_{5}+\:^{22}C_{4} \right)+\:^{23}C_{4}+\:^{24}C_{4}+\:^{25}C_{4} = ?$

0%
$^{22}C_5$
0%
$^{27}C_4$
0%
$^{26}C_4$
0%
$^{26}C_5$

Explanation

This problem will depend on the below given equation

$^{n}C_r + ^{n}C_{r-1} = ^{n+1}C_r$

$\left(\:^{22}C_{5}+\:^{22}C_{4} \right)+\:^{23}C_{4}+\:^{24}C_{4}+\:^{25}C_{4}$

$= \left( \:^{23}C_{5}+\:^{23}C_{4} \right)+\:^{24}C_{4}+\:^{25}C_{4}$

$= \left( \:^{24}C_{5}+\:^{24}C_{4} \right)+\:^{25}C_{4}$

$=\:^{25}C_{5}+\:^{25}C_{4}$

$=\:^{26}C_{5}$

Observe the following Lists

List I	List II
$A)^{ { n } }C_{ r }+^{ n }C_{ r-1 }=$	$1)^{ n+1 }P_{ r }$
$B) ^{ \dfrac {^{n} P_{ r } }{^{n} P_{ r-1 } } = }$	$2)\displaystyle \frac { n-r+1 }{ r }$
$C) ^{n}P_r+r\ {}^{n}P_{r -1 }=$	$3) { n-r+1 }$
$D) { \displaystyle \frac {^{n} C_r }{^{n} C_{ r-1 } } = }$	$4)n+r-1$
	$5)^{(n+1)}C_{ r }$

0%
A-5, B-3, C-2, D-1
0%
A-5, B-3, C-1, D-2
0%
A-2, B-4, C-3, D-1
0%
A-5, B-4, C-2, D-1

Explanation

$^{n}C_{r}+^{n}C_{r-1}=\cfrac{n!}{(n-r)!r!}+\cfrac{n!}{(n-r+1)!(r-1)!}$

$=\cfrac{n!}{(n-r+1)!r!}(n-r+1+r)$

$=\cfrac{(n+1)!}{(n-r+1)!r!}$

$=^{n+1}C_r$

$\cfrac{^{n}P_{r}}{^{n}P_{r-1}}=\cfrac{n!(n-r+1)!}{n!(n-r)!}$

$=n-r+1$

$^nP_r+r. ^nP_{r-1}=\cfrac{n!}{(n-r)!}+r.\cfrac{n!}{(n-r+1)!}$

$=\cfrac{n!}{(n-r+1)}(n-r+1+r)$

$=\cfrac{(n+1)!}{(n-r+1)!}$

$=^{n+1}P_r$

$\cfrac{^nC_r}{^nC_{r-1}} = \cfrac{n!(n-r+1)!(r-1)!}{(n-r)!r!\times n!}$

$=\cfrac{n-r+1}{r}$

$n={ ^{ m }{ C } }_{ 2 }$ , then the value of

${ ^{ n }{ C } }_{ 2 }$ is given by

0%
${ ^{ m+1 }{ C } }_{ 4 }$
0%
${ ^{ m-1 }{ C } }_{ 4 }$
0%
${ ^{ m+2 }{ C } }_{ 4 }$
0%
none of these

Explanation

We have

$n={ _{ }^{ m }{ c } }_{ 2 }^{ } = \displaystyle\frac { m\left( m-1 \right) }{ 2 }$

${ _{ }^{ n }{ c } }_{ 2 }^{ }=\displaystyle\frac { n\left( n-1 \right) }{ 2 }$

$\Rightarrow ^{n}C_2=\dfrac { m\left( m-1 \right) }{ 4 } \left[ \displaystyle\frac { m\left( m-1 \right) }{ 2 } -1 \right] =\displaystyle\frac { 1 }{ 8 } m\left( m-1 \right) \left( { m }^{ 2 }-m-2 \right)$

$\Rightarrow ^{n}C_{2}=\displaystyle\frac { 1 }{ 8 } m\left( m-1 \right) \left( m-2 \right) \left( m+1 \right)$

$\Rightarrow ^nC_2=3\times \displaystyle\frac { 1 }{ 24 } \left( m+1 \right) m\left( m-1 \right) \left( m-2 \right) =3\times { _{ }^{ m+3 }{ c } }_{ 4 }^{ }$

$3.^{ (x+1) }C_{ 2 }+^{ 2 }P_{ 2 }.x=4.^{x}P_{ 2 },x\in N$ , then

$x=$

0%
$2$
0%
$4$
0%
$5$
0%
$3$

Explanation

Given,

$3.^{x+1}C_{2}+^{2}P_{2}x=4.^{x}P_{2}$

$3.\dfrac{(x+1)x}{2}+2x=4x(x-1)$

$3(x+1) +4 = 8(x -1)$

$5x=15$

$\therefore x=3$

Number of points having position vector

$a\hat{i}+b\hat{j}+c\hat{k},$ where

$a,b,c\in \left\{1,2,3,4,5\right\}$ such that

$2^a+3^b+5^c$ is divisible by

$4$ is

0%
$140$
0%
$70$
0%
$100$
0%
none of these

Explanation

$4m={ 2 }^{ a }+{ 3 }^{ b }+{ 5 }^{ c }={ 2 }^{ a }+{ \left( 4-1 \right) }^{ b }+{ \left( 4+1 \right) }^{ c }$

$4m=4k+{ \left( -1 \right) }^{ b }+{ \left( 1 \right) }^{ c }$

Thus if

$a=1$ and

$b$ is even then

$c$ is any number, and

$a\neq 1,b=$ odd, then

$c$ is any number.

Thus required number

$=1\times 2\times 5\times 4\times 3\times 5=70$

Nine hundred distinct

$n$ digit numbers are to be formed using only the three digits

$2,5$ and

$7$ . the smallest value of

$n$ for which this is possible is

0%
$4$
0%
$8$
0%
$7$
0%
$9$

Explanation

Using 2,5 and 7 with repetition, each place of n digits number can be chosen in 3 ways. So total of n digit number is equal to

$=3\times 3\times 3\times ....\times 3\left( n\quad times \right) ={ 3 }^{ n }$

According to the question,

${ 3 }^{ n }\ge 900\Rightarrow { 3 }^{ n-2 }\ge 100\Rightarrow n-2\ge 5\Rightarrow n\ge 7$

The sum

$\sum_{i=0}^{m}\binom{10}{i}\binom{20}{m-i}$ where

$\binom{p}{q} = 0$ ; If (p<q) is maximum when m is

0%
5
0%
10
0%
15
0%
20

Explanation

To find

$\sum_{i=0}^{m}\binom{10}{i}\binom{20}{m-i}$

$\sum_{i=0}^{m}\binom{10}{i}\binom{20}{m-i}=\sum_{i=0}^{m} { }^{10}C_{i}\times { }^{20}C_{m-i}$

$= { }^{10}C_{0}\times { }^{20}C_{m} + { }^{10}C_{1}\times { }^{20}C_{m-1} + { }^{10}C_{2}\times { }^{20}C_{m-2}+\cdot \cdot \cdot+ { }^{10}C_{m}$

$={ }^{30}C_{m}$ ....... Using the above concept

$\therefore \sum_{i=0}^{m}\binom{10}{i}\binom{20}{m-i}={ }^{30}C_{m}$

We know that,

${ }^{n}C_{r}$ is maximum when

$r=\dfrac{n}{2}$

$\therefore { }^{30}C_{m}$ is maximum when

$m=\dfrac{30}{2}=15$

Hence, option C is correct.

Arrange the following values of n in ascending order.

$A: ^nP_5= ^nP_6 \Rightarrow n =$

$B: ^nP_{12}= ^nP_8 \Rightarrow n =$

$C: ^nC_{(n-3)}= 10 \Rightarrow n =$

$D: ^{(n+1)}P_5: ^nP_6 =1:2 \Rightarrow n =$

0%
CABD
0%
CADB
0%
ACBD
0%
DBAC

Explanation

A :

$\displaystyle ^nP_5=^nP_6$

$\Rightarrow \cfrac{n!}{(n-5)!}=\cfrac{n!}{(n-6)!}$

$\Rightarrow n-5=1$

$\Rightarrow n = 6$

B :

$^nP_{12}=^nP_8$

$\Rightarrow n = 20$

$C: ^{n}C_{n-3}=^{n}C_{3}=10$

$\Rightarrow n=5$

$D : \cfrac{(n+1)!}{(n-4)!}\times \cfrac{(n-6)!}{n!}=\cfrac{1}{2}$

$\Rightarrow \cfrac{n+1}{(n-4)(n-5)}=\cfrac{1}{2}$

$\Rightarrow n=9$

Hence, the ascending order of

$n$ is

$=$ CADB

The number of values of

$r$ satisfying the equation

$^{39}C_{3r-1} - ^{39}C_{r^2} = ^{39}C_{r^2-1} - ^{39}C_{3r}$ is:

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$\displaystyle ^{39}C_{3r-1} - ^{39}C_{r^2} = ^{39}C_{r^2-1} - ^{39}C_{3r}$

$\displaystyle ^{39}C_{3r-1}+\, ^{39}C_{3r}=\, ^{39}C_{r\, ^{2}-1}+\, ^{39}C_{r\, ^{2}}$

$\displaystyle ^{40}C_{3r} = ^{40}C_{r^2}$

$\Rightarrow r^2=3r$ or

$3r=40-r^2$

$\Rightarrow r=0, r = 3, or \ r=5,-8$

$... (r=0,-8$ is not possible.)

There are

$n$ different books and

$p$ copies of each in a library. The number of ways in which one or more than one book can be selected is:

0%
$p^n + 1$
0%
${(p+1)}^n - 1$
0%
${(p+1)}^n - p$
0%
$p^n$

Explanation

Number of ways in which one or more objects can be selected out of $S_{1}$ alike objects of one kind, $S_{2}$ alike objects of second kind and $S_{3}$ alike objects of third kind
= $\left (S_{1}+1\right)$ $\left (S_{2}+1\right)$ $\left (S_{3}+1\right) - 1$

The above formula can be generalized as follows.

Number of ways in which one or more objects can be selected out of $S_{1}$ alike objects of one kind, $S_{2}$ alike objects of second kind , $S_{3}$ alike objects of third kind and so on ... $S_{n}$ alike objects of $n^{th}$ kind
$\left(S_{1}+1\right)$ $\left(S_{2}+1\right)$ $\left(S_{3}+1\right)$ ............ $\left(S_{n}+1\right) - 1$

In our case it is $\left(p+1\right)^{n}-1$

The number of quadratic expressions with the coefficients drawn from the set

$( 0, 1, 2, 3 )$ is:

0%
$27$
0%
$36$
0%
$48$
0%
$64$

Explanation

$The\quad number\quad of\quad quadratic\quad expressions\quad with\quad the\quad coefficients\quad drawn\quad from\quad the\quad \\set\quad (0,1,2,3)\quad is\\ Let\quad the\quad Quadratic\quad Expression\quad is\quad ax^{ 2 }+bx+c=0\quad \quad where\quad a\neq 0\\ We\quad have\quad \\ When\quad a=1\quad \quad \quad b=(0,1,2,3)\quad \quad c=(0,1,2,3)\quad \quad \quad Number\quad of\quad Cases=4\times 4=16\\ When\quad a=2\quad \quad \quad b=(0,1,2,3)\quad \quad c=(0,1,2,3)\quad \quad \quad Number\quad of\quad Cases=4\times 4=16\\ When\quad a=3\quad \quad \quad b=(0,1,2,3)\quad \quad c=(0,1,2,3)\quad \quad \quad Number\quad of\quad Cases=4\times 4=16\\ So\quad total\quad Cases\quad =16+16+16=48$

$^nC_0+^{n+1}C_1+ ^{n+2}C_2+ ....+ ^{n+r}C_r =$

0%
$^{n+r+1}C_r$
0%
$^{n+r-1}C_r$
0%
$n$
0%
$^{n+r+2}C_r$

Explanation

$\:^{n}C_{0}+\:^{n+1}C_{1}+...\:^{n+r}C_{r}$

$=(\:^{n}C_{0}+\:^{n+1}C_{0}+\:^{n+1}C_{1}+...\:^{n+r}C_{r})-\:^{n+1}C_{0}$

$=(\:^{n}C_{0}+\:^{n+2}C_{1}+\:^{n+2}C_{2}+...\:^{n+r}C_{r})-\:^{n+1}C_{0}$ ....

$[\because {}^{n+1}C_{0}+{}^{n+1}C_{1}={}^{n+2}C_{1}]$

$=(\:^{n}C_{0}+\:^{n+3}C_{2}+\:^{n+3}C_{3}+...\:^{n+r}C_{r})-\:^{n+1}C_{0}$ ...

$[\because {}^{n+2}C_{1}+{}^{n+2}C_{2}={}^{n+3}C_{2}]$
:
:

$=(\:^{n}C_{0}+\:^{n+r}C_{r-1}+\:^{n+r}C_{r})-\:^{n+1}C_{0}$

$=(\:^{n}C_{0}+\:^{n+r+1}C_{r})-\:^{n+1}C_{0}$ .....

$[\because {}^{n+r}C_{r-1}+{}^{n+r}C_{r}={}^{n+r+1}C_{r}]$

$=1+\:^{n+r+1}C_{r}-1$

$=\:^{n+r+1}C_{r}$ .

$^nC_{r+1} + 2 ^nC_r + ^nC_{r-1} =$

0%
$^{n+1}C_{r}$
0%
$^{n+2}C_{r}$
0%
$^{n+2}C_{r+1}$
0%
$^{n+2}C_{r+2}$

Explanation

$^nC_{r+1} + 2 ^nC_r + ^nC_{r-1}$

$= (^nC_{r+1} + ^nC_r )+ (^nC_r+^nC_{r-1})$

$=^{n+1}C_{r+1}+^{n+1}C_{r} =^{n+2}C_{r+1}$

$n$ is an integer between

$0$ and

$21$ , then the minimum value of

$n! (21-n)$ ! is

0%
$9! 2!$
0%
$10! 11!$
0%
$20!$
0%
$21!$

Explanation

${n}$ !

$(21-{n})!=21$ !

$\displaystyle \frac{n!(21-n)!}{21!}=\frac{(21)!}{21C_{n}}$
For minimum value of,

$\displaystyle \frac{(21)!}{21C_{n}}$ ,

$n$ should be

$10$

$\therefore \dfrac{(21)!}{^{21}C_{10}}=\dfrac{21!}{21!}\times 10!\times 11!=10!11!$

$n$ and

$r$ are integers such that

$1 \le r \le n$ , then

$n.^{n-1}C_{r - 1} =$

0%
$^nC_r$
0%
$n^nC_r$
0%
$r^nC_r$
0%
$(n-1)^nC_r$

Explanation

$n.^{n-1}C_{r - 1} = n.\cfrac{(n-1)!}{(n-r)!(r-1)!}$

$=r.\cfrac{n!}{(n-r)!r!}$

$=r. ^nC_r$

$r > 1$ , then

$\displaystyle \frac{^nP_r}{^nC_r}$ is

0%
is an integer
0%
may be fraction
0%
is an odd number
0%
an even number

Explanation

$\displaystyle \frac{^nP_r}{^nC_r} =\frac{\displaystyle \frac{n!}{(n-r)!}}{\displaystyle \frac{n!}{(n-r)!r!}} =r! =$ even number since

$r>1$

$5$ balls of different colours are to be kept in

$3$ boxes of different sizes. Each box can hold all five balls. Number of ways in which the balls can be kept in the boxes so that no box remain empty is

0%
$60$
0%
$90$
0%
$150$
0%
$200$

Explanation

Arrangement of balls in three boxes:

Let boxes be

$A,B,C$

Case I:

$A,B,C: 1,1,3$ Total

$3$ cases possible.

Total ways for this case:

$3\times\ ^5C_3\ ^2C_1\ ^1C_1 = 3\times10\times2 = 60$

Case II:

$A,B,C: 1,2,2$ Total

$3$ cases possible

Total ways for this case:

$3\times\ ^5C_2\ ^3C_2\ ^1C_1 = 3\times10\times3 = 90$

Total ways:

$60+90 = 150$

The number of solutions of

$^{61}C_{n + 1} = ^{61}C_{2n - 1}$ is

0%
3
0%
1
0%
2
0%
4

Explanation

Given,

$^{61}C_{n + 1} = ^{61}C_{2n - 1}$

$\mathrm{n}+1=2\mathrm{n}-1\\ or\ n+1=61-(2n-1)$

$=>\mathrm{n}=2\ or\ n=\dfrac{61}{3}$

$\therefore \ {n}$ has one solution as n can not be fraction.

$^{(k-1)}C_{(k-1)}+^{k}C_{(k-1)}+^{(k+1)}C_{(k-1)}+ .......+ ^{(n+k-2)}C_{(k-1)}$

$=?$

0%
$^{(n+k)}C_k$
0%
$^{(n+k+1)}C_k$
0%
$^{(n+k)}C_{k-1}$
0%
$^{(n+k-1)}C_k$

Explanation

$\underline{^{k-1}C_{k}+^{k}C_{k-1}}+....+^{n+k-2}C_{k-1}$

$= \underline{ ^{k+1}C_k+^{k+1}C_{k-1}}+....+^{n+k-2}C_{k-1}$

$= ^{k+2}C_{k}+...............+^{n+k-2}C_{k-1} =^{n+k-1}C_{k}$

$^nP_r$ and

$^nC_r$ are equal when:

0%
$n = r$
0%
$n = r + 1$
0%
$r = 1$
0%
$n = r - 1$

Explanation

Given,

$\:^{n}P_{r}=\:^{n}C_{r}$

$\dfrac{n!}{(n-r)!}=\dfrac{n!}{(n-r)!.r!}$

$r!=1$

$r=1$

$10(^nC_2) = 3(^{n+1}C_3)$ , then

$n$

$=$

0%
8
0%
9
0%
10
0%
11

Explanation

Given,

$10(^nC_2) = 3(^{n+1}C_3)$

$10\displaystyle \frac{n!}{2!(n-2)!}=3\frac{(n+1)!}{3!(n-2)!}$

$\Rightarrow 10=({n}+1)$

$\Rightarrow {n}=9$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect and Reason is correct

Explanation

$\dfrac {^nC_{n-r}}{^nC_{n+r}}=\dfrac {r+1}{n-r} \Rightarrow 2n-5r=3$ ....(1)

$\dfrac {^nC_{r-1}}{^nC_r}=\dfrac {r}{n-(n-1)}=\Rightarrow 3n-10r=-3$ ....(2)

Solving (1) and (2), we get

$r=3$

$^nC_2=^3C_2=3$

Thus statement (1) is false and statement (2) is true.

$\displaystyle \frac{n_{C_{r}}}{n_{C_{r-1}}}=$

0%
$\displaystyle \frac{n+r+1}{r}$
0%
$\displaystyle \frac{n-r+1}{r}$
0%
$\displaystyle \frac{n-r-1}{r}$
0%
$\displaystyle \frac{n+r-1}{r}$

Explanation

The value of

$\dfrac{ ^{n}C_{r}}{^{n}C_{r-1}}$

$=\dfrac{n!(n-(r-1))!(r-1)!}{n!(n-r)!r!}$

$=\dfrac{(n+1-r)!(r-1)!}{r!(n-r)!}$

$=\dfrac{(n+1-r)}{r}$
Hence answer is Option B

In the next world cup of cricket, there will be

$12$ teams, divided equally into two groups. Teams of each group will play a match against each other. From each group

$3$ top teams will qualify for the next round. In this round, each team will play against other once. Four top teams from this round, where each team will play against the other three. Two top teams of this round will goto the final round, where they will play the best of three matches. The minimum number of matches in the next world cup will be

0%
$54$
0%
$53$
0%
$38$
0%
$55$

Explanation

No. of matches in first round

$=^6C_{2}+^6C_{2}$
No. of matches in next round

$=^{6}C_{2}$
No. of matches in Semifinal round

$=^4C_{2}$
So required number of matches

$^6C_{2}+^6C_{2}+^6C_{2}+^4C_{2}+2=53$ (minimum

$2$ matches must be played in best of

$3$ )

$^nC_{r} = 10, ^nC_{r+1}=45$ then,

$r$ equals

0%
1
0%
2
0%
3
0%
4

Explanation

Let

$\displaystyle{}^n{C_r} = 10 \Rightarrow \frac{{n!}}{{r!\left( {n - r} \right)!}} = 10$ and

$\displaystyle{}^n{C_{r + 1}} = 45 \Rightarrow \frac{{n!}}{{{r+1}!\left( {n - r-1} \right)!}} = 45$
Hence we get,

$\displaystyle \frac{{{}^n{C_{r + 1}}}}{{{}^n{C_r}}} = \frac{{\frac{{n!}}{{\left( {r + 1} \right)!\left( {n - r - 1} \right)!}}}}{{\frac{{n!}}{{r!\left( {n - r} \right)!}}}} = \frac{{n - r}}{{r + 1}} = \frac{{45}}{{10}} = \frac{9}{2}$
Now we will eliminate the options.
Check that if

$r=2$ or

$r=4$ then

$n$ can not be integer.
For

$r=3$ , check that

${{}^5{C_3}}=10$ , but

${{}^5{C_4}}\ne45$ .
Hence the only possible value for

$r$ is 1 and

$n$ is 10.

The least value of

$n$ so that

$^nC_6 + ^nC_7 > ^{n+1} C_6$ is

0%
13
0%
12
0%
11
0%
10

Explanation

$^nC_6 + ^nC_7 > ^{n+1} C_6$

$^{n+1}C_{7}> ^{n+1}C_{6}\Rightarrow n+1> 7+6$

$=>{n}>12$

$=>{n}=13,14,\ \ldots$

$\therefore {n}=13$

The least positive integral value of

$x$ which satisfies the in equality

$^{10}C_{x-1} >^{10}C_x$ is

0%
$7$
0%
$8$
0%
$9$
0%
$10$

Explanation

For above expression to be defined we must have,

$10\geq x-1\Rightarrow x\leq 11$
And

$10\geq x\Rightarrow x\leq 10$
Given inequation is

$^{10}C_{x-1}> 2.^{10}C_{x}$

$\Rightarrow 1\ge 2\cdot \dfrac{^{10}C_x}{^{10}C_{x-1}}$

$\Rightarrow 1>2.\ \dfrac{10-x+1}x,\because \dfrac{^nC_{r-1}}{^nC_r}=\dfrac{n-r+1}{r}$

$\Rightarrow x> 2(11-x)\Rightarrow 3x>22$

$\Rightarrow \mathrm{x}>22/3$
Using above we get

$x\in(22/3,10]$
But since

$^nC_r$ is only defined for non-negative integral values of

$n$ and

$r$

$\Rightarrow \mathrm{x}=8, 9$
Hence, least positive value is

$8$ .

The number of positive integers satisfying the inequality

$^{(n+1)}C_{n-2}- ^{n+1}C_{n-1}\le$

$100$ is

0%
9
0%
8
0%
5
0%
7

Explanation

$\displaystyle \frac{(n+1)n(n-1)}{6}-\frac{(n+1)n}{2}\leq 100$

$(n+1)n(n-4)\leq 600$
Also

$n-2\geq 0= n\geq 2$
and

$n-1\geq 0= n\geq 1$

$\Rightarrow n\geq 2$

$\therefore$ Possible values are

$\mathrm{n}=2,3,4,5,6,7,8,9$

Let x.y.z

$=$ 105 where

$x, y, z \in N$ . Then number of ordered triplets (x, y, z) satisfying the given equation is

0%
15
0%
27
0%
6
0%
33

Explanation

$xyz=105=3^{1}5^{1}7^{1}$

$3'$ has

$3$ options (i.e whether be in

$x,y$ or

$z$ )

same for

$5$ and

$7$

$\therefore$ Total number of ways =

$3^{3}=27$

$P_n$ denotes the product of all the coefficients in the expansion of

${(1+x)}^n$ and

$9! P_{(n+1)} =10\ ^{9}P_n$ . Then

$n =$

0%
$10$
0%
$9$
0%
$19$
0%
$11$

Explanation

Here,

$9!. P_{(n+1)} =10^{9}.P_n$

$\Rightarrow 9![C_{0}C_{1}\ldots..C_{n+1}]$

$=10^{9}[C_{0}.C_{1}\ldots\ldots C_{n}]$

$\Rightarrow(n+1)_{C_{0}}[\displaystyle \frac{(n+1)_{C_{1}}}{nC_{0}}.\frac{(n+1)_{C_{2}}}{nC_{1}}\ldots..\frac{(n+1)_{C_{n+1}}}{nC_{n}}]=\frac{10^{9}}{9!}$

$\Rightarrow[\displaystyle \frac{n+1}{1}\cdot\frac{n+1}{2}\cdots\cdots\cdots\cdot\frac{n+1}{n+1}]=\frac{10^{9}}{9!}$

$\Rightarrow\displaystyle \frac{(n+1)^{n+1}}{(n+1)!}=\frac{10^{9}}{9!}$

$\Rightarrow n=9$

Let

$\mathrm{S}=\{1,2,3,4\}$ . The total number of unordered pairs of disjoint subsets of

$\mathrm{S}$ is equal to

0%
25
0%
34
0%
42
0%
41

Explanation

Given,

$S=\left \{1, 2, 3, 4\right \}$

Each element can be put in

$3$ ways either in subsets or we don't put in any subset.

Total number of unordered pairs of disjoint subsets,

$\displaystyle =\frac {3^n+1}{2}=\frac {3^4+1}{2}=41$

There are

$4$ candidates for a Natural science scholarship,

$2$ for a Classical and

$6$ for a Mathematical scholarship,then find No. of ways these scholarships can be awarded is,

0%
$48$
0%
$12$
0%
$24$
0%
$8$

Explanation

Natural science scholarship can be awarded to anyone of the four candidates. So, there are

$4$ ways of awarding natural science scholarships.

Similarly, mathematical and classical scholarships can be awarded in

$2$ and

$6$ ways respectively.

Hence, by fundamental principle of multiplication, number of ways of awarding these scholarships

$=4\times 2\times 6=48$

Hence option-

$A$

Given:

$\dfrac {20!}{18!}=380$

0%
True
0%
False
0%
Either
0%
Neither

Explanation

$\cfrac { 20! }{ 18! } =?\\ 20!=20\times 19\times 18\times ........3\times 2\times 1\\ =20\times 19\times (18!)\\ \cfrac { 20! }{ 18! } =\cfrac { 20\times 19\times (18!) }{ 18! } =20\times 19=380$

Hence the value matches so the equation is true.

In a combination, the ordering of the selected objects is immaterial whereas in a permutation, the ordering is essential.

0%
True
0%
False
0%
Either
0%
Neither

$\sum_{r=0}^{n} r^{2} . ^{n}C_{r}p^{r}q^{n-r} , where \ \ p + q = 1,$ is simplified to:

0%
$npq + n^{2}p^{2}$
0%
$n^{2}p^{2}q^{2} + np$
0%
$np(p + q)$
0%
$\frac{p(q+1)}{2}$

Explanation

Given:

$\sum _{ r=0 }^{ n }{ { r }^{ 2 }{ x }_{ r }{ p }^{ r }{ q }^{ n-r } }$

$=\sum _{ r=0 }^{ n }{ { r }^{ 2 }\times \dfrac { n! }{ r!(n-r)! } { p }^{ r }{ q }^{ n-r } }$

$=\sum _{ r=0 }^{ n }{ \dfrac { n!\times r\times r }{ (r-1)!(n-r)!r } { p }^{ r }{ q }^{ n-r }+{ n }^{ 2 }p^{ n } }$

$=\sum _{ r=0 }^{ n }{ \dfrac { n(n-1)! }{ (r-1)!(n-r)! } p\times (r{ p }^{ r-1 })q\times { q }^{ n-r-1 }+{ n }^{ 2 }p^{ n } }$

$=npq\sum _{ r=0 }^{ n }{ _{ \quad }^{ n-1 }{ { C }_{ r-1 } }p\times (r{ p }^{ r-1 }){ q }^{ n-r-1 }+{ n }^{ 2 }p^{ n } }$

$=npq(^{ n-1 }{ { C }_{ 0 } }p^{ 0 }+{ q }^{ n-2 })+(^{ n-2 }{ { C }_{ 1 }2 }p^{ 1 }+{ q }^{ n-3 })+(^{ n-1 }{ { C }_{ n-1 } }p^{ n-2 }{ q }^{ 0 }+{ n }^{ 2 }p^{ n })$

$=npq+{ n }^{ 2 }p^{ n }$

Which among the following is/are correct?

0%
If an operations can be performed in 'm' different ways and a second operation can be performed in 'n' different ways, then both of these operations can be performed in $'m\times n'$ ways together.
0%
The number of arrangements of n different objects taken all at a times is n!.
0%
The number of permutations of n different things taken r at a time, when each thing may be repeated any number of times is n.
0%
The number of circular permutations of 'n' different things taken all at a time is $\frac {1}{2}(n-1)!$ , if clockwise and anticlockwise orders are taken as different.

Explanation

$A)$ if an operation can be performed in

$m$ ways and the second operation can be performed in

$n$ ways,then the number of ways in which these two operations can be performed are

$m\times n$ ways,

because for each operation in the first there are

$n$ ways for the second one ,so these

$n$ operations add up

$m$ times,

which is

$n\times m$

$B)$ the number of arrangements of

$n$ things taken all at a time is

$n!$ because,

We need to

$n$ things in

$n$ places,so there are

$n$ ways to fill a thing in the first gap and

$n-1$ ways for the second gap,proceeding like this we get

$(n)(n-1)(n-2)...(3)(2)(1)=n!$

$C)$ the number of permutations of n different things taken r at a time, when each thing may be repeated any number of times is greater than or equal to n and not n.

$D)$ The number of circular permutations of 'n' different things taken all at a time is

$\dfrac{1}{2} (n-1)!$ if clockwise and anticlockwise orders are same and not different.

Which among the following is/are not correct?

0%
$^6C_2+ \ ^6C_1= \ ^7C_2$
0%
$^6C_1+ \ ^6C_2= \ ^6C_2$
0%
$^6C_2+ \ ^6C_1= \ ^7C_1$
0%
$^6C_2- \ ^6C_1= \ ^7C_2$

Explanation

We know that

$C(n,r) + C(n,r+1)=C(n+1,r+1)$

Applying the above rule, only option (1) follows and hence options (2), (3) and (4) are correct answers.

Two series of a question booklet for an aptitude test are to be given to twelve students. In how many ways can the students be placed in two rows of six each so that there should be no identical series side by side and that the students sitting one behind the other should have the same series?

0%
$2 \times ^{12}C_6 \times(6!)^2$
0%
6! x 6!
0%
7! x 7!
0%
2 x 6!

Explanation

(B)

$6! \times 6!$

How many different signals can be transmitted by arranging 3 red, 2 yellow and 2 green flags on a pole? [Assume that all the 7 flags are used to transmit a signal].

0%
210
0%
215
0%
220
0%
225

Explanation

Here

$n=3+2+2=7$

$p_{1}=3, p_{2}=2$ and

$p_{3}=2$

$\therefore$ Required number of different signals

$=\frac{n!}{p_{1}!p_{1}!p_{1}!}=\frac{7!}{3!2!1!}\frac{7.6.5.4}{2.2}$

$=7.6.5=210$

A five digit number divisible by

$3$ is to be formed using the numerals

$0, 1, 2, 3, 4$ and

$5$ without repetition. The total number of ways in which this can be done is:

0%
$211$
0%
$216$
0%
$221$
0%
$311$

Explanation

All permutations formed with

$1, 2, 3, 4, 5$ (sum

$=15$ ) will be divisible by

$3$ .
There are

$5!=120$ such permutations.

Such numbers can also be formed using

$0$ and

$1, 2, 3, 4, 5$ .

There are

$4\times 4!$ such numbers, i.e.

$96$ .

(Factors of

$4$ for four positions of

$0$ and

$4!$ for different permutations of these four numbers)

$\therefore$ Total of such numbers

$=120+96=216$

How many strings of letters can possibly by formed using the above rules such that the third letter of the string is e?

0%
8
0%
9
0%
10
0%
11

Explanation

1st letter	2nd letter	3rd letter	Total No of ways
a,i,o,u	m	e	$4\times1\times 1=4$
a,e,i,o,u	n	e	$5\times1\times 1=5$
e	p	e	$1\times 1\times1=1$

Total no of ways =

$10$

How may strings of letters can possibly be formed using the above rules?

0%
40
0%
45
0%
30
0%
35

How many numbers greater than a million can be formed with the digits

$2, 3, 0, 3, 4, 2, 3$ ?

0%
$360$
0%
$366$
0%
$356$
0%
$370$

Explanation

A number greater than a million has

$7$ places, and thus all the

$7$ given digits are to be used.

But

$2$ is repeated twice and

$3$ is repeated thrice.

$\therefore$ Total number of ways of arranging these

$7$ digits amongst themselves

$=\dfrac{7!}{2!3!}=420$
But numbers beginning with zero are no more seven digits numbers

Thus, we have to reject those numbers which begin with zero and such numbers are

$\dfrac{6!}{2!3!}=60$
Hence, the required number of arrangements

$=420-60=360$

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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