MCQ Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 5

$\displaystyle \begin{pmatrix} 35 \\ 6 \end{pmatrix}+\sum_{r=0}^{10}\begin{pmatrix} 45-r \\ 5 \end{pmatrix}=\begin{pmatrix} x \\ y \end{pmatrix}$
then

$x-y$ is equal to:

0%
$39$
0%
$29$
0%
$52$
0%
$40$

Explanation

Expanding we get

$\:^{35}C_{6}+\:^{35}C_{5}+\:^{36}C_{5}+\:^{37}C_{5}+\:^{38}C_{5}+\:^{39}C_{5}+\:^{40}C_{5}+\:^{41}C_{5}+\:^{42}C_{5}+\:^{43}C_{5}+\:^{44}C_{5}+\:^{45}C_{5}$
Now

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

$=\:^{n+1}C_{r+1}$
Applying, this, and simplifying, we get

$\:^{46}C_{6}$
Hence,

$46-6$

$=40$

$\displaystyle \begin{pmatrix} 47 \\ 4 \end{pmatrix}$ +

$\displaystyle \sum_{j=1}^{5}$

$\displaystyle \begin{pmatrix} 52-j \\ 3 \end{pmatrix}$ =

$\displaystyle \begin{pmatrix} x \\ y \end{pmatrix}$ then

$\displaystyle \frac{x}{y}$ =

0%
11
0%
12
0%
13
0%
14

Explanation

Expanding, we get

$\:^{47}C_{4}+\:^{47}C_{3}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}+\:^{51}C_{3}+\:^{52}C_{3}$
Applying

$\:^{n}C_{r}+\:^{n}C_{r+1}=\:^{n+1}C_{r+1}$ , we get

$[\:^{47}C_{4}+\:^{47}C_{3}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}+\:^{51}C_{3}]$

$=+\:^{48}C_{4}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}+\:^{51}C_{3}]$
:
:

$\:^{51}C_{3}+\:^{51}C_{4}$

$=\:^{52}C_{4}$ .
Hence

$x=52$ and

$y=4$
Hence,

$\dfrac {x}{y}=\dfrac{52}{4}$

$=13$

$^8C_r = ^8C_{r+2}$ , then the value of

$^rC_2$ is:

0%
8
0%
3
0%
5
0%
2

Explanation

Since

${ C }_{ r }^{ n }={ C }_{ n-r }^{ n }$

$\Longrightarrow r=8-(r+2)$

Hence

$r=3$

$^3C_2=3$

$^{50}C_{11}+^{50}C_{12}+^{51}C_{13}-^{52}C_{13}=$

0%
$^{52}C_{14}$
0%
$0$
0%
$^{53}C_{13}$
0%
none of these

Explanation

We know that

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

Application of the above formula gives us

$\:^{50}C_{11}+\:^{50}C_{12}$

$=\:^{51}C_{12}$

$\rightarrow \:^{51}C_{12}+\:^{51}C_{13}$

$=\:^{52}C_{13}$

$\rightarrow \:^{52}C_{13}-\:^{52}C_{13}$

$=0$ .

The value of

$^{95}C_4+\displaystyle \sum_{j=1}^5 {\;}^{100-j}C_3$ is

0%
$^{95}C_5$
0%
$^{100}C_4$
0%
$^{99}C_4$
0%
$^{100}C_5$

Explanation

$^{95}C_4+\displaystyle \sum_{j=1}^5 {\;}^{100-j}C_3$

$= ^{95}C_4 + ^{99}C_3 + ^{98}C_3 + ^{97}C_3 + ^{96}C_3 + ^{95}C_3$

$= (^{95}C_3 + ^{95}C_4) + ^{96}C_3 + ^{97}C_3 + ^{98}C_3 + ^{99}C_3$

$= (^{96}C_4 + ^{96}C_3) + ^{97}C_3 + ^{98}C_3 + ^{99}C_3$ (

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

$= (^{97}C_4 + ^{97}C_3) + ^{98}C_3 + ^{99}C_3$ (

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

$= (^{98}C_4 + ^{98}C_3) + ^{99}C_3$ (

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

$= ^{99}C_4 + ^{99}C_3 = ^{100}C_4$ (

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

$^nC_3 + ^nC_4 > ^{n+1}C_3$ , then

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

Explanation

$\:^{n}C_{3}+\:^{n}C_{4}$

$=\:^{n+1}C_{4}$

$=\:^{n+1}C_{n-3}$
Now

$\:^{n+1}C_{n-3}>\:^{n+1}C_{3}$
Hence

$n-3>3$

$n>6$ .

$'n'$ is an integer between

$0$ and

$21$ , then the minimum value of

$n!\left( 21-n \right) !$ is

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

Explanation

To find the minimum value of

$\displaystyle n!\left( 21-n \right) !=21!\frac { n!\left( 21-n \right) ! }{ 21! } =21!\frac { 1 }{ ^{ 21 }{ { C }_{ n } } } =\frac { 21! }{ ^{ 21 }{ { C }_{ n } } }$

For minimum value,

$\displaystyle ^{ 21 }{ { C }_{ n } }$ is maximum.

Maximum value of

$\displaystyle ^{ 21 }{ { C }_{ n } }=\frac { ^{ 21 }{ { C }_{ 21-1 } } }{ 2 } =^{ 21 }{ { C }_{ 10 } }$

$\therefore$ Minimum value

$\displaystyle =\frac { 21! }{ ^{ 21 }{ { C }_{ 10 } } } =\frac { 21! }{ 21! } \times 10!11!=10!11!$

$\displaystyle \sum_{r=0}^m {\;}^{n+r}C_n=$

0%
$^{n+m+1}C_{n+1}$
0%
$^{n+m+2}C_{n}$
0%
$^{n+m+3}C_{n+1}$
0%
none of these

Explanation

Given

$\sum _{ r=0 }^{ m } {^{ n+r }C_{ n }}$

$=^{ n }C_{ n }+^{ n+1 }C_{ n }+^{ n+2 }C_{ n }+^{ n+3 }C_{ n }.....^{ n+m }C_{ n }$

$=^{ n+1 }C_{ n+1 }+^{ n+1 }C_{ n }+^{ n+2 }C_{ n }+^{ n+3 }C_{ n }.....^{ n+m }C_{ n }$ (

$\because ^{ n }C_{ n }=^{ n+1 }C_{ n+1 }$ )

$=^{ n+2 }C_{ n+1 }+^{ n+2 }C_{ n }+^{ n+3 }C_{ n }.....^{ n+m }C_{ n }$
(

$\because ^nC_r+^nC_{r-1}=^{n+1}C_r$ )

$=^{ n+3 }C_{ n+1 }+^{ n+3 }C_{ n }+......^{ n+m }C_{ n }$
Continuing this ,

$=^{ n+m }C_{ n+1 }+^{ n+m }C_{ n }$

$=^{ n+m+1 }C_{ n+1 }$

$2\times$

$^nC_5 = 9\times$

$^{n-2}C_5$ , then the value of n will be:

0%
7
0%
10
0%
9
0%
5

Explanation

Given,

$2(\:^{n}C_{5})=9(\:^{n-2}C_{5})$

$2.\dfrac{n!}{5!.(n-5)!}=9.\dfrac{(n-2)!}{(n-7)!.5!}$

$\dfrac{2(n-1)(n)}{(n-6)(n-5)}=9$

$2n^{2}-2n=9(n^{2}-11n+30)$

$7n^{2}-97n+270=0$
Hence

$n=10$ and

$n=\dfrac{27}{7}$
Since

$n\epsilon N$ .
Hence

$n=10$ .

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

0%
3
0%
4
0%
5
0%
8

Explanation

Since,

$^nC_r = ^nC_{n-r}$

Either

$\Longrightarrow 3r=r+3$ or

$3r=15-(r+3)$

Hence,

$r=\dfrac { 3 }{ 2 }$ or

$r=3$

Since r is an integer, therefore

$r=3$

Let

$A=(x|x$ is a prime number and

$x<300$ > the number of different rational numbers, whose numerator and denominator belong to

$A$ is:

0%
91
0%
84
0%
106
0%
None of these

Explanation

Set A contains 10 elements. 2 different numbers for numerator and denominator from these can be obtained in

$10\times 9=90$ ways and each permutation will form a unique rational number different from one. In addition, one will be formed when numerator and denominator are same. So required number of numbers

$90+1=91$

If n and r are two positive integers such that

$n\geq r$ , then

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

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

Explanation

The value of

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

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

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

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

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

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

$(^{15}C_r + ^{15}C_{r-1}) (^{15}C_{15-r} + ^{15}C_{16-r}) = (^{16}C_{13})^2$ , then the value of

$r$ is

0%
$r=2$
0%
$r=3$
0%
$r=4$
0%
none of these

Explanation

$(^{ 15 }C_{ r }+^{ 15 }C_{ r-1 })(^{ 15 }C_{ 15-r }+^{ 15 }C_{ 16-r })=(^{ 16 }C_{ 13 })^{ 2 }$

$^{ 16 }C_{ r }(^{ 15 }C_{ r }+^{ 15 }C_{ r-1 })=(^{ 16 }C_{ 13 })^{ 2 }$ (

$\because ^nC_r+^nC_{r-1}=^{n+1}C_r$ )

$^{ 16 }C_{ r }(^{ 16 }C_{ r })=(^{ 16 }C_{ 3 })^{ 2 }$ (

$\because ^nC_r=^nC_{n-r}$ )

$(^{ 16 }C_{ r })^{ 2 }=(^{ 16 }C_{ 3 })^{ 2 }$

$\Rightarrow r=3$

In a game called 'odd man out',

$m (m > 2)$ persons toss a coin to determine who will buy refreshment for the entire group. A person who gets an outcome different from that of the rest of the members of the group is called the odd man out. The probability that there is a loser in any game is

0%
$1/2 m$
0%
$m/2 ^{m-1}$
0%
$2/m$
0%
none of these

Explanation

Let us represent the person as

${ P }_{ 1 }$ ,

${ P }_{ 2 }$ ,

${ P }_{ 3 }$ ,

${ P m }.$

For each person we have 2 outcome,

So, total outcomes

$= { 2 }^{ m }$

Favorable outcomes:

$1.$ One gets tail rest

$\rightarrow$ Head

$2.$ One get head test

$\rightarrow$ Tail

No. of ways

$m\times 1$ (

$m=$ because in people).

Also for second condition we have

$m$ ways.

Total favorable ways

$=\dfrac { 2m }{ { 20 }^{ m } } ={ m }/{ { 2 }^{ m-1 } }$

Hence, answer is

${ m }/{ { 2 }^{ m-1 } }.$

Three players play a total of

$9$ games. In each game, one person wins and the other two lose; the winner gets

$2$ points and the losers lose

$1$ each. The number of ways in which they can play all the

$9$ games and finish each with a zero score is

0%
$84$
0%
$1680$
0%
$7056$
0%
$0$

Explanation

Each player need to win

$3$ matches and lose

$6$ matches to get a zero score.

Out of

$9$ games

$3$ should be selected by first player(games which he will win)

$=^9C_3$

Out of

$6$ games 3 should be selected by second player

$={}^6C_3$

Out of

$3$ games

$3$ should be selected by third player

$={}^3C_3$

Hence the answer is

$={}^9C_3\times^6C_3\times^3C_3=1680$

Let

$S= \left\{1,2,3,.......n\right\}$ and

$A =\left\{(a, b) \left. \right | 1 \geq a, b \geq n\right\} = S \times S\:$ . A subset

$B$ of

$A$ is said to be a good subset if

$(x, x) \in B$ for every

$x \in S.$ Then the number of good subsets of

$A$ is

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

Explanation

The number of element in

$A$ =

$n\times{n}$

The number of element of A having

$(x,x)$ for all

$x\in S$ =

$n$

Total remaining element =

$(n\times{n})-n$

Since Subset

$B$ must have

$(x,x)$ for all

$x\in S$ =

$n$ so it may have any number of remaining element.

Number of ways of selecting elements from remaining element =

${2}^{(n\times{n})-n}$

$4.^nC_6 = 33.^{n-3}C_3$ then

$n$ is equal to

0%
$9$
0%
$10$
0%
$11$
0%
none of these

Explanation

$4.\:^{n}C_{6}=33.(\:^{n-3}C_{3})$

$4.(\dfrac{n!}{(n-6)!.6!})=33.(\dfrac{(n-3)!}{3!.(n-6)!}$

$4(n(n-1)(n-2))=33(4\times5\times6)$

$n(n-1)(n-2)=11\times3\times5\times6=11.10.9$

Hence the real root that we get is

$n=11$ .
Hence

$n=11$ .

There are 8 buses running from Kota to Jaipur and 10 buses running from Jaipur to Delhi. In how many ways a Person can travel from Kota to Delhi via Jaipur by bus?

0%
80
0%
8
0%
10
0%
160

Explanation

Let

$E_1$ be the event of travelling from Kota to Jaipur &

$E_2$ be the event of travelling from Jaipur to Delhi by the person.

$E_1$ can happen in 8 ways and

$E_2$ can happen in 10 ways.
Since both the events

$E_1$ and

$E_2$ are to be happened 12 in order, simultaneously,the number of ways

$=8\times10=80$

Forty teams play a tournament. Each of them plays with every other team just once. Each game result is a win for one team. If each team has a

$50\%$ chance of winnings each game, the number of ways such that at the end of the tournament, every team has won a different number of games is:

0%
$\dfrac{1}{780}$
0%
$40!2^{780}$
0%
$40! 3^{780}$
0%
None of these

One mapping is selected at randon from all mappings of the set

$S = \left\{ 1,2,3, ....., n \right\}$ into itself. If the probability that mapping is one-one is

$3/32$ then the value of

$n$ is

0%
$2$
0%
$3$
0%
$4$
0%
None of these

Explanation

Total number of possible mapping

$=n\times n\times n.....n$

$n$ times

$=n^n$ times

Favorable cases

$:n\times(n-1)(n-2)....$

Favorable cases are are found by removing the possibility mapping of the number by which other number is mapped initially.

So,

$P=\dfrac{n!}{n^n}$

$=\dfrac{3\times8}{16\times2\times8}$

$\Rightarrow$

$n=4.$

Hence, the answer is

$4.$

The sum of the series

$\displaystyle \sum_{r=0}^{10} {\;}^{20}C_r$ is

0%
$2^{20}$
0%
$2^{19}$
0%
$2^{19}+\displaystyle \frac {1}{2}^{20}C_{10}$
0%
$2^{19}-\displaystyle \frac {1}{2}^{20}C_{10}$

Explanation

Simplifying the expression we get

$\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+...\:^{20}C_{10}$ ...(i)
Now

$(1+x)^{20}=\:^{20}C_{0}+\:^{20}C_{1}x+\:^{20}C_{2}x^{2}+...\:^{20}C_{20}x^{n}$
Substituting x=1, we get

$2^{20}=\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+...\:^{20}C_{20}$
Now

$\:^{20}C_{r}=\:^{20}C_{n-r}$
Hence

$2^{20}=2[\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+...\:^{20}C_{9}]+\:^{20}C_{10}$
Now

$2^{19}-\dfrac{\:^{20}C_{10}}{2}=\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+...\:^{20}C_{9}$
Adding

$\:^{20}C_{10}$ on both sides we get

$2^{19}-\dfrac{\:^{20}C_{10}}{2}+\:^{20}C_{10}=\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+...\:^{20}C_{9}+\:^{20}C_{10}$

$2^{19}+\dfrac{\:^{20}C_{10}}{2}=\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+...\:^{20}C_{9}+\:^{20}C_{10}$

$\displaystyle^{n}P_{r}= 360$ and

$^ {n}C_{r}=15$ then find the value of r

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

Explanation

$^nP_r = \dfrac { n! }{ (n-r)! } = 360$ -- (1)

$^nC_r = \dfrac { n! }{ (r!)(n-r)! } = 15$ -- (2)

Dividing eqn 1 by 2, we get

$=> r ! = \dfrac {360}{15} = 24$

$=> r ! = 4 \times 2 \times 3 \times 1 = 4 !$

So,

$r = 4$

The value of

$\displaystyle ^{ 19 }C_{ 18 }+^{ 19 }C_{ 17 }$

0%
$1200$
0%
$2000$
0%
$190$
0%
None of these

Explanation

$\displaystyle ^{ n }C_{ r }=\frac { n! }{ \left( n-r \right) !r! }$

$\displaystyle ^{ 19 }C_{ 18 }+^{ 19 }C_{ 17 }=\frac { 19! }{ 18! } +\frac { 19! }{ 17!2! }$

$\displaystyle =19+\frac { 19\times 18 }{ 2 }$

$\displaystyle =19+171$

$\displaystyle =190$

$\displaystyle ^{n}C_{r}=^{n}P_{r}$ then r can be____

0%
0
0%
1
0%
3
0%
Either(1) or (2)

Explanation

Given,

$^nC_r=^nP_r$

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

$=> r ! = 1$

$=> r = 0$ or

$1$ as

$0 ! = 1; 1 ! = 1$

There are

$5$ roads leading to a town from a village. The number of different ways in which a villager can go to the town and return back, is

0%
$25$
0%
$20$
0%
$10$
0%
$5$

Explanation

$\textbf{Step:-1 Calculate total number of outcomes}$

$\text{Given that,there are 5 roads leading to a town from village}$

$\text{The number of ways that the villager can go and return to this village =}$

$5\times5=25$

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

$C_0 + 2C_1+ 3 C_2 + ...... (n + 1) C_n$ will be

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

Explanation

$C_0 + 2 C_1 + 3C_2 + ..... + (n+1)C_n$

$= ^nc_0 + 2^nC_1 + 3^nC_2 + ..... + (n + 1) ^nC_n$

$(^n C_0 + ^nC_1 + ^nC_2 + ..... + ^nC_n) + (^nC_1 + 2 ^nC_2 + 3^nC_3 + ..... n. ^nC_n)$

$= \displaystyle (1 +1)^n + n + n (n - 1) + 3 \frac{n (n -1)(n - 2)}{3} + .... + n$

$= 2^n + n \displaystyle \left [ 1 + (n - 1) + \frac{(n - 1) (n - 2)}{2} + ...... + 1 \right ]$

$= 2^n + n [1 + 1]^{n - 1} = 2^n + n.2^{n -1}= (n + 2). 2^{n - 1}$

Find the value of

$\displaystyle ^{10}C_{10}$

0%
$2$
0%
$1$
0%
$4$
0%
$6$

Explanation

We have

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

So,

$^{10}C_{10} = \dfrac { 10! }{ (10!) (10-10)! } =\dfrac { 1! }{ 0! } = 1$ Since

$1 ! = 0 ! = 1$

In how many ways can 8 books be distributed among 5 students if each student is eligible for any number of books?

0%
$40$
0%
$\displaystyle 5^{8}$
0%
$360$
0%
$\displaystyle 8^{5}$

Explanation

First book can be distributed to student in 5 ways, 2nd book again by 5 ways.

So, using same logic the total ways

$=5\times5\times5.......8$ times

$=5^8$

Hence, the answer is

$5^8.$

$\displaystyle ^{5}C_{4}=$ ___

0%
$4$
0%
$5$
0%
$20$
0%
$24$

Explanation

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

So,

$^5C_4 = \dfrac { 5! }{ (4!) (5-4)! } =\dfrac { 5! }{ 4! } = \dfrac {5 \times 4 !}{4 ! } = 5$

Different calenders for the month of February are made so as to serve for all the coming years. The number of such calenders is

0%
$7$
0%
$2$
0%
$14$
0%
None of these

$\displaystyle ^{3}C_{r}$ the value of r can be ___

0%
$2$
0%
$4$
0%
$6$
0%
All of these

Explanation

A combination is a "one or more elements selected from a set in any order

So,

${3}_{{ C }_{ r }}$ means

$r$ elements are selected from

$3$ elements

This is possible only when

$r \leq 3$
In the given options,

$r = 2$ is the correct answer.

A three-digit code for certain locks uses the digits

$0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ according to the following constraints. The first digit cannot be

$0$ or

$1$ , the second digit must be

$0$ or

$1$ , and the second and third digits cannot both be

$0$ in the same code. How many different codes are possible?

0%
$144$
0%
$152$
0%
$160$
0%
$168$
0%
$176$

Explanation

The first digit can be fill in

$8$ ways. For

$2nd$ digit it can be one of

$(0,1)$

Now, assume second digit as

$1$ the

$3rd$ digit can take

$10$ values.

$\therefore$ Number of codes

$=8\times1\times9=72$

$\therefore$ total

$80+72=152$

Hence, the answer is

$152.$

In how many ways can 3 diamond cards be drawn simultaneously from a pack of cards?

0%
$78$
0%
$1716$
0%
$286$
0%
$13$

Explanation

Pack of cards

$52$

Diamond cards are

$13$

To select

$3$ cards

Total number of choices =

$^{ 13 }C_{ 3 }$

$= \dfrac { 13\times 12\times 11 }{ 3\times 2\times 1 } \\ = 143\times 2\\ = 286$

Hence

$286$ ways are possible and is a correct answer.

A researcher plans to identify each participant in a

$174$ certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are

$12$ participants,and each participant is to receive a different code?

0%
$4$
0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

The number of single letter code possible are

$n$ and no of distinct codes

$={^n{C}_{2}}$

So,

${^n{C}_{2}}+n\ge 12$

$\dfrac{n(n-1)}{2}+n\ge 12$

$n(n+1)\ge 24$

$n$ min

$=5$ as least number is asked.

Hence the answer is

$5.$

The diagram above shows the various paths along which a mouse can travel from point X, where it is released, to point Y, where it is rewarded with a food pellet. How many different paths from X to Y can the mouse take if it goes directly from X to Y without retracing any point along a path?

0%
$6$
0%
$7$
0%
$12$
0%
$14$
0%
$17$

Explanation

We can calculate the total number of ways by using the basic principle for counting by multiplication.

This we do by multiplying the total possible routes between every two nodes in the path form X to Y. (different node are named in the image )

$\rightarrow$ A

$\rightarrow$ C and X

$\rightarrow$ B

$\rightarrow$ C ------------- 2 ways

C to D has to be covered to move ahead.

$\rightarrow$ E

$\rightarrow$ G and D

$\rightarrow$ F

$\rightarrow$ G ------------- 2 ways

G to H has to be covered to move ahead.

$\rightarrow$ I

$\rightarrow$ J and H

$\rightarrow$ K

$\rightarrow$ J ------------- 3 ways

J to Y must be moved to reach Y.

So, total ways =

$2 \times 2 \times 2 \times 3 = 12$ (option C)

Mario's Pizza has 2 choices of crust: deep dish and thin-and-crispy. The restaurant also has a choice of 5 toppings: tomatoes, sausage, peppers, onions, and pepperoni. Finally, Mario's offers every pizza in extra cheese as well as regular. If Linda's volleyball team decides to order a pizza with 4 toppings, how many different choices do the teammates have at Mario's Pizza?

0%
$24$
0%
$32$
0%
$28$
0%
$20$

Explanation

20 choices: Consider the toppings first. Model the toppings with the "word" YYYYN, in which four of the toppings are on the pizza and one is not. The number of anagrams for this "word" is:

$\dfrac{5!}{4!} = 5$

If each of these pizzas can also be offered in 2 choices of crust, there are

$5 \times 2 = 10$ pizzas. The same logic applies for extra cheese and regular:

$10 \times 2 = 20.$

From the consecutive integers -10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

0%
$\displaystyle { \left( -10 \right) }^{ 20 }$
0%
$\displaystyle { \left( -10 \right) }^{ 10 }$
0%
$\displaystyle 0$
0%
$\displaystyle { -\left( 10 \right) }^{ 19 }$
0%
$\displaystyle { -\left( 10 \right) }^{ 20 }$

Explanation

For least value we take

$10;19$ times

$-10$ once

So value

$10^{19}\times-10$

$=(-10)^{20}$

Hence the answer is

$(-10)^{20}.$

$\displaystyle ^{ 16 }{ C }_{ r }=^{ 16 }{ C }_{ r+1 }$ , then the value of

$\displaystyle ^{ r }{ P }_{ r-3 }$ is

$\displaystyle$

0%
31
0%
120
0%
210
0%
840
0%
No option is correct

Explanation

$\displaystyle ^{ 16 }{ C }_{ r }=^{ 16 }{ C }_{ r+1 }$

$\displaystyle \Rightarrow \quad ^{ 16 }{ C }_{ 16-r }=^{ 16 }{ C }_{ r+1 }\quad \left( \because \quad ^{ n }{ C }_{ n-r } \right)$

$\displaystyle \Rightarrow \quad 16-r=r+1$

$\displaystyle \Rightarrow \quad 2r=15$

$\displaystyle \Rightarrow \quad r=7.5$

Which is not possible, since r should be an integer.

Four speakers will address a meeting where speaker

$Q$ will always speak after

$P$ . Then, the number of ways in which the order of speakers can be prepared is

0%
$256$
0%
$128$
0%
$24$
0%
$12$

Explanation

Four speakers will address the meeting in

$4!$ ways

$= 24$ different ways in which half number of cases will be such that

$P$ speaks before

$Q$ and half number of case will be such that

$P$ speaks after

$Q$

$\therefore$ Required number of ways

$=\dfrac {24}{2} = 12$

$_{ }^{ n }{ { P }_{ r } }=30240$ and

$_{ }^{ n }{ { C }_{ r } }=252$ , then the ordered pair

$\left( n,r \right)$ is equal to

0%
$\left( 12,6 \right)$
0%
$\left( 10,5 \right)$
0%
$\left( 9,4 \right)$
0%
$\left( 16,7 \right)$

Explanation

Given that,

$_{ }^{ n }{ { P }_{ r } }=30240$ and

$_{ }^{ n }{ { C }_{ r } }=252$

$\Rightarrow \dfrac { n! }{ \left( n-r \right) ! } =30240$ and

$\dfrac { n! }{ \left( n-r \right) !r! } =252$

$\Rightarrow r!=\dfrac { 30240 }{ 252 } =120$

$\Rightarrow r=5$

$\therefore \dfrac { n! }{ \left( n-5 \right) ! } =30240$

$\Rightarrow n\left( n-1 \right) \left( n-2 \right) \left( n-3 \right) \left( n-4 \right) =30240$

$\Rightarrow n\left( n-1 \right) \left( n-2 \right) \left( n-3 \right) \left( n-4 \right) =10\left( 10-1 \right) \left( 10-2 \right) \left( 10-3 \right) \left( 10-4 \right)$

$\Rightarrow n=10$
Hence, required ordered pair is

$\left( 10,5 \right)$

$^{n}C_{12} = ^{n}C_{6}$ , then

$^{n}C_{2}$ is equal to:

0%
$72$
0%
$153$
0%
$306$
0%
$2556$

Explanation

Given

$^{n}C_{12} = ^{n}C_{6}$
or

$^{n}C_{n - 12} = ^{n}C_{6}$

$\Rightarrow n - 12 = 6$

$\Rightarrow n = 18$

$\therefore ^{n}C_{2} = ^{18}C_{2} = \dfrac {18\times 17}{2\times 1}$

$= 153$

$^n C_{r-1}\, =\, 330,\, ^nC_r\, =\, 462,\, ^nC_{r+1}\, =\, 462\, \Rightarrow\, r\, =$

0%
3
0%
4
0%
5
0%
6

Explanation

$^nC_{r-1}=330$ ,

$^nC_r=^nC_{r+1}=462$

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

$\Rightarrow (r+1)!(n-(r+1))!=(r!)\,(n-r)!$

$\Rightarrow r+1=n-r$

$n=2r+1$

$\dfrac{n-r+1}{r}=\dfrac{462}{330}$

$= 330n-330r+330=462r$

$=330n+330=792r$

$=(n+1)330=792r$

$=(2r+2)330=792r$

$=660r+660=792r$

$=660=132r$

$\Rightarrow r=\dfrac{660}{132}$

$\rightarrow r=5$

A vehicle registration number consists of

$2$ letters of English alphabet followed by

$4$ digits, where the first digit is not zero. Then, the total number of vehicles with distinct registration numbers is

0%
$26^{2} \times 10^{4}$
0%
$^{26}P_{2} \times\ ^{10}P_{4}$
0%
$^{26}P_{2} \times 9\times\ ^{10}P_{3}$
0%
$26^{2} \times 9\times 10^{3}$

Explanation

The total number of arrangements of

$2$ letters of English alphabet

$= 26\times 26$
The total number of arrangements of

$4$ digits number in which first digit is not zero

$= 9\times 10\times 10\times 10$

$\therefore$ The total number of vehicles with distinct registration number

$= 26\times 26\times 9\times 10\times 10\times 10$

$= 26^{2} \times 9\times 10^{3}$

$a=99^{50}+100^{50}$ and

$b=101^{50}$ , then :

0%
$a < b$
0%
$a=b$
0%
$a > b$
0%
$a-b=100^{49}$

Explanation

$a = 99^{50} + 100^{50}$

$b = 101^{50} = (100 + 1)^{50}$

$b = ^{100} C_{0} 100^{50} + ^{100}C_{1} 100^{49} + ... + ^{100}C_{100} 100^{0} = 2 \times {100}^{50} + ^{100}C_2 100^{48} + .... + ^{100}C_{100} 100^{0}$

$b = 100^{50} + 100^{50} + ^{100}C_2 100^{48} + .... + ^{100}C_{100} 100^{0}$

$b = a - 99^{50} +100^{50} + ^{100}C_2 100^{48} + .... + ^{100}C_{100} 100^{0}$

$\because \left(100^{50} - 99^{50}\right) > 1$ and

$^{100}C_2 100^{48} + .... + ^{100}C_{100} 100^{0} >1$

Hence,

$b > a$

If n is an integer with

$0\le n \le 11$ then the minimum value of

$n!(11-n)!$ is attained when a value of n =

0%
11
0%
5
0%
7
0%
9

Explanation

Let

$y=n!(11-n)!$

Consider

$x= ^{11}C_n=\dfrac{11!}{n!(11-n)!}$

For maximum value of

$x$ we must have

$n=6$ or

$n=5$

Thus

$x_{max}=^{11}C_6=\dfrac{11!}{5!6!}=\dfrac{11!}{y_{min}}$

$\Rightarrow y_{min}=5!6!$ i.e.

$n=6$ or

$n=5$

Kathy is scheduling the first four periods of her school day. She needs to fill those periods with calculus, art, literature, and physics, and each of these courses is offered during each of the first four periods. Calculate the total number of different schedules Kathy can choose from.

0%
$1$
0%
$4$
0%
$12$
0%
$24$
0%
$120$

Explanation

Number of options of selecting a course for

$1^{st}$ period is

$4$ (there are

$4$ courses )

Number of options of selecting a course for

$2^{nd}$ period is

$3$

Number of options of selecting a course for

$3^{rd}$ period is

$2$

Therefore total number of different schedules is

$4 \times 3 \times 2 \times 1 = 24$ .

$^{(n-1)}C_3+^{(n-1)}C_4 > ^nC_3$ , then the minimum value of

$n$ is:

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

Explanation

$^{(n-1)}C_3+^{(n-1)}C_4 > ^nC_3$ ,

Use

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

$\Rightarrow\dfrac{n!}{(n-4)!4!}>\dfrac{n!}{(n-3)!3!}$

$\Rightarrow \dfrac{1}{4}>\dfrac{1}{n-3}$

$\Rightarrow n-3>4\Rightarrow n>7$

Smallest value of

$n=8$

The number of words that can be formed out of the letters of the word "ARTICLE" so that the vowels occupy even places is

0%
$574$
0%
$36$
0%
$754$
0%
$144$

Explanation

There are total

$7$ words then there will be

$7$ positions out of

$3$ are even place like

$2,4,6$ and we have

$3$ vowels

$A,I,E$

The vowels can be arranged in the even place in

$3!$ places and the rest can be arranged in

$4!$

$\therefore$ Required number of ways

$3!\times4!=144$

If k is odd then

$\displaystyle ^{ k }{ C }_{ r }$ , is maximum for r equal to

0%
$\displaystyle \frac { 1 }{ 2 } \left( k-1 \right)$
0%
$\displaystyle \frac { 1 }{ 2 } \left( k+1 \right)$
0%
$\displaystyle k-1$
0%
$\displaystyle k$

$X, Y, Z, D, E$ and

$F$ are

$6$ distinct points on the circumference of a circle, find the number of different chords which can be drawn using any

$2$ of the

$6$ points.

0%
$6$
0%
$12$
0%
$15$
0%
$30$
0%
$36$

Explanation

To draw a chord we choose any two points from given 6 points .

The answer required is

$^6C_2=15$

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 5 - 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 5 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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