MCQ Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 12

When six fair coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads?

0%
$25$
0%
$41$
0%
$22$
0%
$42$

Explanation

Total number of coins $=6$

(1) '0' heads $=1$ chance $\left( ^{ 6 }{ C }_{ 0 } \right)$

(2) '1' heads $=6$ chance $\left( ^{ 6 }{ C }_{ 1 } \right)$

(3) '2' heads $=15$ chance $\left( ^{ 6 }{ C }_{ 2 } \right)$

(4) '3' heads $=20$ chance $\left( ^{ 6 }{ C }_{ 3 } \right)$

Hence total outcome $=42$

The number of ways in which four letters can be selected from the word 'DEGREE' is

0%
$7$
0%
$6$
0%
$\dfrac{6!}{3!}$
0%
None of these

Explanation

There are

$6$ letters in the word 'DEGREE', namely

$3\text{E's}$ and

$\text{D, G, R}$ . Four letters out of six can be selected in the following ways
(i)

$3$ like letters and

$1$ different, viz.

$\text{EEE + D, G, R}$
(ii)

$2$ like letters and

$2$ different, viz.

$\text{EEE}\ +$ any two of

$\text{D, G, R}$ .
(iii) All different letters, viz.

$\text{EDGR}$ .
So, the total number of ways

$=^3C_1 + ^3C_2 + 1 = 3 + 3 + 1 = 7$

$^{15}C_{3r} = ^{15}C_{r+3}$ , then find the value of

$r$ :

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

Explanation

$^{ 15 }{ C }_{ 3r }=^{ 15 }{ C }_{ r+3 }$

We Know

$^{ 15 }{ C }_{ 3r }=^{ 15 }{ C }_{ r+3 }\\ ^{ n }{ C }_{ r }=^{ n }{ C }_{ n-r }\\ n=r+(n-r)\\ =3r+r+3\\ =4r+3\\ \Rightarrow 4r+3=15\\ \Rightarrow 4r=12\\ \Rightarrow r=3$

The value of

$\sum^{10}_{r=0}\begin{pmatrix}10\\r\end{pmatrix}\begin{pmatrix}15\\14-r\end{pmatrix}$ is equal to

0%
$^{25}C_{12}$
0%
$^{25}C_{15}$
0%
$^{25}C_{10}$
0%
$^{25}C_{14}$

Two lines intersect at

$O$ . Points

$A_{i}$ and

$B_{i} (i = 1, 2, ...., n)$ are taken on these two lines respectively, the number of triangles that can be drawn with the help of these

$2n + 1$ points is

0%
$n$
0%
$n^{2}$
0%
$n^{3}$
0%
None of these

Explanation

Number of triangles

$=$ Taking any two points in

$B_{1}, B_{2}, ..., B_{n}$ and one point from

$A_{1}, A_{2}, ...., A_{n}$ or taking any two points from

$A_{1}, A_{2}, A_{3}, ....A_{n}$ and one point from

$B_{1}, B_{2}, ..., B_{n}$ or one point

$O$ and one point from

$A_{1}, A_{2} ....A_{n}$ and one point from

$B_{1}, B_{2}, ..., B_{n}$

$\therefore$ Required ways

$= ^{n}C_{2}\times ^{n}C_{1} + ^{n}C_{1}\cdot ^{n}C_{2} + ^{n}C_{1} \times ^{n}C_{1} \times 1$

$= \dfrac {n^{2}(n - 1){2}} + \dfrac {n^{2}(n - 1)}{2} + n^{2}$

$= (n^{2}(n - 1) + n^{2}(n - 1) + 2n^{2})/2$

$= n^{2} [2n - 2 + 2]/2 = n^{3}$

Value of

$\sum _{ k=0 }^{ n }{ { _{ }^{ k }{ C } }_{ n }\sin { \left( kx \right) } \cos { \left( n-k \right) } }$ is

0%
${ 2 }^{ n-1 }\left( \sin { nx } \right)$
0%
${ 2 }^{ n }\sin { \left( nx \right) }$
0%
${ 2 }^{ n }\cos { \left( nx \right) }$
0%
none of these

In a triangle

$ABC$ , the value of the expression

$\displaystyle \sum_{r = 0}^{n}\ ^{n}C_{r}a^{r}.b^{n - r}.\cos (rB - (n - r)A)$ is equal to

0%
$c^{n}$
0%
zero
0%
$a^{n}$
0%
$b^{n}$

Seven person

$P_1,P_2......, P_7$ initially seated at chairs

$C_1,C_2,.....C_7$ respectively.They all left there chairs simultaneously for hand wash. Now in how many ways they can again take seats such that no one sits on his own seat and

$P_1$ , sits on

$C_2$ and

$P_2$ sits on

$C_3$ ?

0%
$52$
0%
$53$
0%
$54$
0%
$55$

$m$ denotes the number of

$5$ digit numbers if each successive digits are in their descending order of magnitude and

$n$ is the corresponding figure. When the digits and in their ascending order of magnitude then

$(m-n)$ has the value

0%
${ _{ }^{ 9 }{ C } }_{ 4 }$
0%
${ _{ }^{ 9 }{ C } }_{ 5 }$
0%
${ _{ }^{ 10 }{ C } }_{ 3 }$
0%
${ _{ }^{ 9 }{ C } }_{ 3 }$

Explanation

${\textbf{Step 1: Find m}}$

${\text{For m,}}$

${\text{First we select any 5 digits from 0,1,2,}}...{\text{,9}}$

${\text{Number of ways = }}{}^{10}{{\text{C}}_5}$

${\text{Now after selection there is only 1 way to arrange these selected digits, i}}{\text{.e}}{\text{., in descending order}}{\text{.}}$

${\text{Therefore m = }}{}^{10}{{\text{C}}_5}\times{\text{ 1 = }}{}^{10}{{\text{C}}_5}$

${\textbf{Step 2: Find n}}$

${\text{For n,First we select any 5 digits from 1,2,}}...{\text{,9}}$

${\text{We can't select zero as first digit because then the number won't be a 5 - digit number}}{\text{.}}$

${\text{Therefore number of ways = }}{}^9{{\text{C}}_5}$

$\Rightarrow {\text{n = }}{}^9{{\text{C}}_5}\times{\text{ 1 = }}{}^9{{\text{C}}_5}$

$\Rightarrow {\text{m - n = }}{}^{10}{{\text{C}}_5}{\text{ - }}{}^9{{\text{C}}_5}$

${\text{We know that,}}{}^n{{\text{C}}_r}{\text{ + }}{}^n{{\text{C}}_{r - 1}}{\text{ = }}{}^{n + 1}{{\text{C}}_r}$

$\Rightarrow {}^{n + 1}{{\text{C}}_r}{\text{ - }}{}^n{{\text{C}}_r}{\text{ = }}{}^n{{\text{C}}_{r - 1}}$

$\Rightarrow {}^{10}{{\text{C}}_5}{\text{ - }}{}^9{{\text{C}}_5}{\text{ = }}{}^9{{\text{C}}_4}$

${\text{Hence, m - n = }}{}^9{{\text{C}}_4}$

${\textbf{Hence, the correct answer is option A}}$

There are

$2$ identical white balls,

$3$ identical red balls and

$4$ green balls of different shades. The number of ways in which they can be arranged in a row so that atleast one ball is separated from the balls of the same colour, is

0%
$6(7! - 4!)$
0%
$7(6! - 4!)$
0%
$8! - 5!$
0%
None

Explanation

These are totally

$9$ balls of which

$2$ are identical of one kind,

$3$ are a like of another kind

$and$

$4$ district ones.

At least one ball of same color separated

$=$ Total

$-$ No ball of same color is separated

Total permutation

$=\dfrac{9!}{2!3!}$

For no ball is separated : we consider all balls of same color as

$1$ entity, so there are

$3$ entities which can be placed in

$3!$ ways.

The white and red balls are identical so they will be placed in

$1$ way whereas green balls are different so they can be placed in

$4!$ ways

$\Rightarrow Req=3!\times 4!$

At least one ball is separated

$=\dfrac{9!}{2!3!}-3!4!$

$=\dfrac{9\times 8\times 7!}{2\times 6}-6\times 4!$

$=6\left(7!\right)-6\left(\times 4!\right)$

$=6\left( 7!-4!\right ).$

Hence, the answer is

$6\left( 7!-4!\right ).$

Two classrooms A and B having capacity of

$25$ and

$(n-25)$ seats respectively.

$A_n$ denotes the number of possible seating arrangements of room

$'A'$ , when 'n' students are to be seated in these rooms, starting from room

$'A'$ which is to be filled up to its capacity. If

$A_n-A_{n-1}=25!(^{49}C_{25})$ then 'n' equals:

0%
$50$
0%
$48$
0%
$49$
0%
$51$

Explanation

Given

$A_n=nC_{25} \cdot 25!$

$A_{n-1}={n-1}C_{25} \cdot 25!$

Hence

$nC_{25} \cdot 25! - (n-1)C_{25} \cdot 25!=25! 49C_{25}$

$\Rightarrow (n-1)C_{25}+(n-1)C_{24}-(n-1)C_{25}=49C_{24}$

$\Rightarrow n-1=49$

$\Rightarrow n=50$

$\displaystyle \sum_{k = 1}^{n = 1} \ ^{n - k}C_{r} = ^{x}C_{y}$ then

0%
$x = n + 1; y = r$
0%
$x = n; y = r + 1$
0%
$x = n; y = r$
0%
$x = n + 1; y = r + 1$

${ \left( 1+x \right) }^{ n }=\sum _{ r=0 }^{ n }{ { a }_{ r }{ x }^{ r } }$ and

${ b }_{ r }=1+\cfrac { { a }_{ r } }{ { a }_{ r-1 } }$ and

$\prod _{ r=1 }^{ n }{ { b }_{ r }=\cfrac { { \left( 101 \right) }^{ 100 } }{ 100! } }$ , then

$n$ equals to:

0%
$99$
0%
$100$
0%
$101$
0%
None of these

Sum of series :

$\sum_\limits{r=1}^n (r^2+1)( r! )$ is _______

0%
(n+1) !
0%
(n+2) ! - 1
0%
n(n + 1) !
0%
None of these

There are infinite, alike, blue, red, green and yellow balls. Find the number of ways to select

$10$ balls.

0%
$286$
0%
$84$
0%
$715$
0%
None of these.

Let

$(1+x)^n=C_0+C_1x+C_2x^2+.....+C_nX^n$ . (where

$C_r=^nC_r$ )
On the basis of given information, answer the following question.

$C_1-3(C_3)+5(C_5)-7(C_7)+$ _______ is?

0%
$n.2^{\left(\dfrac{n-1}{2}\right)}$
0%
$n\left(2^{\dfrac{(n-1)}{2}}\right)\cos \dfrac{n\pi}{4}$
0%
$n(2^{(n-1)})$
0%
$n.2^{\dfrac{(n-1)}{2}}\cos\left(\dfrac{(n-1)\pi}{4}\right)$

Maximum number of common chords of a parabola and a circle can be equal to

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

parabola

$eq^n : x^2 = 4ay ...(1)$

circle

$eq^n : x^2 +y^2 + 2fx + 2gy + h =0 ...(2)$

substituting (1) in(2)

$[y = x^2 / 4a]$

$x^2 + \dfrac{x^2}{(4a)^2}+2fx + 2gy +h = 0$

This is a degree 4 polynomial and has 4 distinct real roots,

so point of interaction b/w circle & parabola

$\Rightarrow 4$

Since common chords can be formed poly any

$2$ points out of

$4$ point of interaction.

$\therefore$ common chords

$= ^{4}C_{2} = \dfrac{4\times 3 }{2 \times 1} = \boxed{6}$

(option C)

In a certain examination paper, there are

$n$ question. For

$j = 1, 2....n$ , there are

$2^{n-j}$ students who answered

$j$ or more questions wrongly. If the total number of wrong answers is

$4095$ , then the value of

$n$ is:

0%
$12$
0%
$11$
0%
$10$
0%
$9$

How many

$10-digit$ numbers can be formed by using the digits

$1$ and

$2$ ?

0%
$^{10}{P}_{2}$
0%
$^{10}{C}_{2}$
0%
${2}^{10}$
0%
$10\ !$

Explanation

Each place of a ten digit number can be fixed by any of the two digits. So, the number of ways to form a ten digit number is ${2^{10}}$ .

The value of

$\displaystyle\sum^{10}_{r=0}(r)$

$^{20}C_r$ is equal to?

0%
$20(2^{18}+{^{19}}C_{10})$
0%
$10(2^{18}+{^{19}}C_{10})$
0%
$20(2^{18}+{^{19}}C_{11})$
0%
$10(2^{18}+{^{19}}C_{11})$

Explanation

$\begin{matrix} \sum _{ r=0 }^{ 10 }(r){ \, ^{ 20 } }{ C_{ r } } \\ \Rightarrow 0{ +^{ 20 } }{ C_{ 1 } }+{ 2^{ 20 } }{ C_{ 2 } }+{ 3^{ 20 } }{ C_{ 3 } }+............+{ 10^{ 20 } }{ C_{ 10 } } \\ \Rightarrow 20+20\times 19+\dfrac { { 20\times 19\times 18 } }{ 2 } +.........+10\times \dfrac { { 20! } }{ { 10!\times 10! } } \\ \Rightarrow 20\left[ { 1+19+\dfrac { { 19\times 18 } }{ 2 } +\dfrac { { 19\times 18\times 17 } }{ 6 } +...........+\dfrac { { 10\times 19! } }{ { 10!\times 10! } } } \right] \\ \Rightarrow 20\left[ { 1+19+\dfrac { { 19\times 18 } }{ 2 } +\dfrac { { 19\times 18\times 17 } }{ 6 } +.........+\dfrac { { 19! } }{ { 9!\times 10! } } } \right] \\ \, \, \, \, \, \, 20\left[ { ^{ 19 }{ C_{ 0 } }{ +^{ 19 } }{ C_{ 1 } }{ +^{ 19 } }{ C_{ 2 } }{ +^{ 19 } }{ C_{ 3 } }+..........{ +^{ 19 } }{ C_{ 10 } } } \right] \, \, \, \, Ans. \\ \end{matrix}$

What is the value of

$^nC_n$ ?

0%
zero
0%
$1$
0%
$n$
0%
$n!$

Let

$S$ be the set of

$6$ digits numbers

$a_{1},a_{2},a_{3},a_{4},a_{5},a_{6}$ (all digits distinct) where

$a_{1}>a_{2}>a_{3}<a_{4}>a_{5}<a_{6}$ . Then

$n(S)$ is equal to

0%
$210$
0%
$2100$
0%
$4200$
0%
$420$

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

0%
140
0%
196
0%
280
0%
346

Explanation

There would be two cases,

(1) When 4 is selected from first five and rest 6 from remaining 8.

This is done in

$^{5}C_{4}\times ^{8}C_{6}$ ways.

(2) When all 5 are selected from first five and rest five from remaining 8.

This can be done in

$^{5}C_{5}\times ^{8}C_{5}$ ways.

$\therefore$ A total of

$^{5}C_{4}\times ^{8}C_{6}+^{5}C_{5}\times ^{8}C_{5}$

$=5\times \cfrac{8!}{6!2!}+\cfrac{8!}{3!5!}$

$=\cfrac{5\times 8\times 7}{2}+\cfrac{8\times 7\times 8}{3\times 2}$

$=140+56=196\,ways$

A library has

$'a'$ copies of one book,

$'b'$ copies of each of two books,

$'c'$ copies of each of three books, and single copy each of

$'d'$ . The total number of ways in which these books can be arranged in a row is

0%
$\dfrac{(a+b+c+d)!}{a!b!c!}$
0%
$\dfrac{(a+2b+3c+d)!}{a!(b!)^{2}(c!)^{3}}$
0%
$\dfrac{(a+2b+3c+d)!}{a!b!c!}$
0%
none of these

$S_n=\displaystyle\sum^{n}_{r=0}\dfrac{1}{^{n}C_r}$ and

$t_n=\displaystyle\sum^n_{r=0}\dfrac{r}{^{n}C_r}$ , then

$\dfrac{t_n}{S_n}$ is equal to?

0%
$\dfrac{1}{2}n$
0%
$\dfrac{1}{2}n-1$
0%
$n-1$
0%
$\dfrac{2n-1}{2}$

$2C_0 + \dfrac{2^2}{2} C_1 + \dfrac{2^3}{3} C_2 +........+ \dfrac{2^{11}}{11} C_{10}$ is equal to

0%
$\dfrac{2^{11} -1}{11}$
0%
$\dfrac{3^{11} -1}{11}$
0%
$\dfrac{3^{11} -1}{12}$
0%
$\dfrac{3^{11} +1}{11}$

The number of 5 letter words formed using letters of word "CALCULUS" is

0%
$280$
0%
$15$
0%
$1110$
0%
$56$

$^{2n}C_{3}:^{n}C_{2}::44:1$ , then the value of

$n$ is

0%
$17$
0%
$6$
0%
$11$
0%
$none\ of\ these$

If m denotes the number of

$5$ digit numbers if each successive digits are in their descending order of magnitude and n is the corresponding figure, when the digits are in their ascending order of magnitude then

$(m-n)$ has the value?

0%
$^{10}C_4$
0%
$^9C_5$
0%
$^{10}C_3$
0%
$^9C_3$

Explanation

$\textbf{Step-1: For descending order}$

$\text{In this case, we can choose 5 numbers from 10 numbers.}$

$\text{This can be done in}$

$^{10}C_5$

$\text{ways.}$

$\text{Now, as there is only one way in which five numbers can be arranged,}$

$m=^{10}C_5.$

$\textbf{Step-2: For ascending order}$

$\text{In this case, we have 9 options to choose from, as here, 0 will be in front and the }$

$\text{number will be 4-digit.}$

$\text{Hence,}$

$n=^{9}C_4$

$\textbf{Step-3: Finding}$

$\mathbf{m-n}$

$m-n=^{ 10}C_5-^9C_4$

$=^9C_5$

$\textbf{Hence, the correct option is (B).}$

$10$ IIT and

$2$ NIT students sit at random in a row, and then number of ways in which exactly

$3$ IIT students sit between

$2$ NIT students is

0%
$16\times 10!$
0%
$15\times 10!$
0%
$10\times 16!$
0%
$None\ of\ these$

${ if}^{ n }{ c }_{ 10 }{ = }^{ n }{ c }_{ 14 }$ then the value of n is equal to

0%
14
0%
24
0%
34
0%
44

${ if }^{ n }{ c }_{ r }{ + }^{ n }{ c }_{ r+1 }={ n+1 }_{ { C }_{ x } }$ then the value of x is equal to

0%
r
0%
r + 1
0%
r - 1
0%
2r

The sum

$^{ n }{ C }_{ 0 }+^{ n }{ C }_{ 1 }+^{ n }{ C }_{ 2 }+.......+^{ n }{ C }_{ n }$ is equal to

0%
$\frac { 2.4.6.........2n }{ n\quad ! }$
0%
${ n }^{ n }$
0%
$n!$
0%
${ 3 }^{ n }$

$\displaystyle\sum^m_{k=1}(k^2+1)k!=1999(2000!)$ , then m is?

0%
$1999$
0%
$2000$
0%
$2001$
0%
$2002$

$n\ \in\ N$ &

$n$ is even, then

$\dfrac {1}{1\ .\ (n-1)\ !}+\dfrac {1}{3\ !(n-3)\ !}+\dfrac {1}{5\ !\ (n-5)\ !}+....+\dfrac {1}{(n-1)\ !\ 1\ !}=$

0%
$2^{n}$
0%
$\dfrac {2^{n-1}}{n\ !}$
0%
$2^{n}n\ !$
0%
$none\ of\ these$

The value of

$\sum _{ r=0 }^{ n }{ \sum _{ s=0 }^{ n }{ (r+s){ C }_{ r }{ C }_{ s } } } ,$ is

0%
$n.2^{2n}$
0%
$n(n-1)2^{n-2}$
0%
$n(n+1)2^n$
0%
$2^{2n}$

$^{ 2017 }{ c }_{ 0 }{ + }^{ 2017 }{ c }_{ 1 }{ + }^{ 2017 }{ c }_{ 2 }+.....{ + }^{ 2017 }{ c }_{ 1008 }={ \lambda }^{ 2 }(\lambda >0),$ then remainder when

$\lambda$ is divided by 33 is

0%
8
0%
13
0%
17
0%
25

Explanation

Given

$^{2017}C_{0}+\, ^{2017}C_{1}+\, ^{2017}C_{2}+….+\, ^{2017}C_{1008}= \lambda^{2}$

$(\lambda > 0)$

$\Rightarrow \displaystyle \sum^{1008}_{k=0}$

$^{2017}C_{k}= \lambda^{2}$

He we know that,

$\displaystyle \sum^{n}_{k=0}$

$^{n}C_{k}=2^{n}$

$\displaystyle \therefore \sum_{k=0}^{2017}$

$^{2017}C_{k}=2^{2017}$

$\Rightarrow \displaystyle \sum_{k=0}^{1008}$

$^{2017}C_k$

$+\sum_{k=1009}^{2017}$

$^{2017}C_{k}=2^{2017}$

$\Rightarrow \displaystyle \sum_{k}^{1018}$

$^{2017}C_{k}+$

$^{2017}C_{1009}$

$+$

$^{2017}C_{1010}+…..+$

$^{2017}C_{2017}=2^{2017}$

$\Rightarrow \displaystyle \sum_{k=0}^{1008}$

$^{2017}C_{k}+$

$^{2017}C_{1008}+$

$^{2017}C_{1007}+…….+$

$^{2017}C_{0}=2^{2017}$

$[using\ ^{m}C_{r}=^{m}C_{m-r}]$

$\Rightarrow \displaystyle \sum_{k=0}^{1008}$

$^{2017}C_{k}$

$+ \displaystyle \sum_{1008}^{k=0}$

$^{2017}C_{k}= 2^{2017}$

$\Rightarrow \displaystyle 2 \sum_{k=0}^{1008}$

$^{2017}C_{k}=2^{2017}$

$\Rightarrow \displaystyle \sum_{k=0}^{1008}$

$^{2017}C_{k}=2^{2016}$

$\Rightarrow\displaystyle \sum_{k=0}^{1008}$

$^{2017}C_{k}= (2^{1008})^{2}$

$\therefore$

$\lambda=2^{1008}$

$\lambda=2^{3}.2^{1005}$

$=2^{3} (2^{5})^{201}$

$=2^{3} (32)^{201}$

$=2^{3} (33-1)^{201}$ [using binomial theorem ]

$=8[^{201}C_{0} (33)^{201}+\, ^{201}C_{1} (33)^{200} (-1)+\, ^{201}C_{2} (33)^{199} (-1)^{2}+…..+\, ^{201}C_{201} (-1)^{201}]$

$=8[(33)^{201}+201 (33)^{200} (-1)+\, ^{201}C_{2} (33)^{199} (-1)^{2}+….+(-1)^{201}]$

$=[8 (33)^{201}+8.201 (300)^{200} (-1)+ 8\times ^{201}C_{2} (33)^{199} (-1)^{2}+….-8]$

When

$\lambda$ is divided by

$33$ then the remainder is

$-8$ which is negative we can also write the remainder as

$33-8=25$

${ if } ^ { n }{ C }_{ 3 }{ + }^{ n }{ C }_{ 4 }>{ n+1 }_{ C_{ 3 } }$ then

0%
n > 6
0%
n > 7
0%
n < 6
0%
none of these

Value of

$\sum _{ r=1 }^{ n }{ \left( \sum _{ m=0 }^{ r }{ { }_{ } } \right) } ^nC_r,^rC_m)$ is equal to

0%
$2^n-1$
0%
$3^n-1$
0%
$3^n-2^n$
0%
$None$ $of$ $these$

$(1+x)^n=C_0+C_1x+C_2x^2+....+C_nX^n$ , then the value of

$C^2_0+\dfrac{C^2_1}{2}+\dfrac{C^2_2}{3}+....+\dfrac{C^2_n}{n+1}$ is equal to?

0%
$\dfrac{(2n-1)!}{\{(n+1)!\}^2}$
0%
$\dfrac{(2n-1)!}{(n+1)!}$
0%
$\dfrac{2n+1)!}{\{(n+1)!\}^2}$
0%
$\dfrac{(2n)!}{(n+1)!}$

$(1+x)^{n}=C_{0}+C_{1}x+C_{2}x^{2}+...+C_{n}x^{n}$ then the value of

$1^{2}C_{1}+2^{2}C_{2}+3^{2}C_{3}+...+n^{2}C_{n}$ is

0%
$n(n+1)2^{n-2}$
0%
$n(n+1)2^{n-1}$
0%
$n(n+1)2^{n}$
0%
None of these

If n is odd natural number, then

$\sum _{ r=0 }^{ n }{ \cfrac { { (-1) }^{ r } }{ ^{ n }{ C }_{ r } } }$ equals

0%
0
0%
1/n
0%
$n/{ 2 }^{ n }$
0%
None of these

There are

$n$ points on a circle. The number of staight lines formed by joining them is equal to

0%
$^{n}C_{2}$
0%
$^{n}P_{2}$
0%
$^{n}C_{2}-1$
0%
$none\ of\ these$

The number of ways

$5$ identical balls can be distributed into

$3$ different boxes so that no box remains empty.

0%
$^{4}C_2$
0%
$6$
0%
$^{4}C_1$
0%
$^{4}C_3$

The number of words which can be made out of the letters of the word

$'MOBILE'$ when consonants always occupy odd places is _______.

0%
$20$
0%
$36$
0%
$30$
0%
$720$

In a set of lottery Tickets

$7$ carry prizes and

$25$ are blank. If three tickets are drawn then the probability to get a prize is?

0%
$\dfrac{{^{7}C_3}}{{^{32}C_3}}$
0%
$\dfrac{{^{25}C_3}}{{^{32}C_3}}$
0%
$1-\dfrac{{^{25}C_3}}{{^{32}C_3}}$
0%
Cannot be decided

The sum of the numbers formed by taking all the digit at a from 0, 2, 3, 4 is

0%
57996
0%
75669
0%
99657
0%
57699

Number of six digit numbers whose sum of the digits is

$49$ are

0%
$^{12}C_{5}$
0%
$6^{10}$
0%
$^{10}C_{5}$
0%
$1200$

The letters of the word 'VICTORY' are arranged in all possible ways, and the words thus obtained are arranged as in a dictionary. Then the rank of given word is

0%
3733
0%
5309
0%
5040
0%
3732

$20$ soldiers are standing in a row and their captain want to send

$7$ out of them for a mission. In how many ways can captain select them such that at least one soldier find the soldier next to him is also selected.

0%
$^{20}C_{7}$
0%
$^{14}C_{7}$
0%
$^{20}C_{7}-^{13}C_{7}$
0%
$^{20}C_{7}-^{14}C_{7}$

Explanation

$\textbf{Step -1: Calculating}$

$\text{Required number of ways= (Total number of ways without any restriction )}$

$\text{- (No. of ways in which no two consecutive soldiers are selected)}$

$\implies \text{Let }x_1\text{ be the number of soldiers selected before the first soldier is selected.}$

$\text{Also , let }x_2,x_3,x_4..x_7 \text{ be the number of soldiers between 1st and 2nd, 2nd and 3rd}$

$\text{ ...6th and 7th respectively}$

$\text{Also let }x_8 \text{ be the number of soldiers after the 7th soldier is selected}$

$\textbf{Step -2: Calculating the number of ways .}$

$\implies x_1+x_2+...+x_8=13$

$x_1,x_8 \geq 0 \text{ and }x_2,x_3..,x_7 \geq 1$

$\implies x_1+y_2+...+y_7+x_8=7$

$\text{Where }y_2,y_3..,y_7\geq 0$

$\text{Number of solution of above equation}$

$\implies ^{8+7-1}C_{8-1}= ^ { 14}C_7$

$\therefore \text{Required number of ways = } ^{20}C_7-^ {14}C_7$

$\textbf{Hence, the required number of ways is }\mathbf{^{20}C_7- ^{14}C_7 .}$

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 12 - 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 12 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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