MCQ Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 8

If the letters of the word "VARUN" are written in all possible ways and then are arranged as in a dictionary, then the rank of the word VARUN is?

0%
$98$
0%
$99$
0%
$100$
0%
$101$

Explanation

${\textbf{Step 1}}\;:\;{\mathbf{Finding}}\;{\mathbf{number}}\;{\mathbf{of}}\;{\mathbf{words}}\;{\mathbf{starting}}\;{\mathbf{with}}\;{\mathbf{the}}\;{\mathbf{alphabets}}\;{\mathbf{in}}\;{\mathbf{the}}\;{\mathbf{given}}\;{\mathbf{word}}$

${\text{No}}.{\text{ of words beginning with A }} = 4! = 4 \times 3 \times 2 = 24$

${\text{No}}.{\text{ of words beginning with N }} = 4! = 4 \times 3 \times 2 = 24$

${\text{No}}.{\text{ of words beginning with R }} = 4! = 4 \times 3 \times 2 = 24$

${\text{No}}.{\text{ of words beginning with U }} = 4! = 4 \times 3 \times 2 = 24$

${\text{So}},{\text{ in total }}96{\text{ words will be formed while beginning with letter A}},{\text{ N}},{\text{ R and U}}.$

${\textbf{Step 2}}\;\;:\;{\mathbf{Finding}}\;{\mathbf{the}}\;{\mathbf{order}}\;{\mathbf{of}}\;{\mathbf{the}}\;{\mathbf{word}}$

${\text{Order of }} 97{\text { th word }} - {\text{ VANRU}}$

${\text{Order of }}98{\text{th word }} - {\text{ VANUR}}$

${\text{Order of }}99{\text{th word }} - {\text{ VARNU}}$

${\text{Order of }}100{\text{th word}} - {\text{ VARUN}}.$

${\mathbf{Therefore}},\;{\mathbf{rank}}\;{\mathbf{of}}\;{\mathbf{word}}\;'{\mathbf{VARUN}}'\;{\textbf{is 100. Option C is correct.}}$

Garlands are formed using 6 red roses and 6 yellow roses of different sizes. The number of arrangements in garland which have red roses and yellow roses come alternately is

0%
$5!\times 6!$
0%
$6! \times 6!$
0%
$\dfrac{5!}{2!}\times 6!$
0%
$2(6!\times 6!)$

The value of

$\sum\limits_{r = 0}^{10} {\left( {\matrix{ {10} \cr r \cr } } \right)} \left( {\matrix{ {15} \cr {14 - r} \cr } } \right)$ is equal to

0%
${}^{25}{C_{12}}$
0%
${}^{25}{C_{15}}$
0%
${}^{25}{C_{10}}$
0%
${}^{25}{C_{11}}$

The product of five consecutive numbers is always divisible by ?

0%
$60$
0%
$12$
0%
$120$
0%
$72$

Explanation

Solution:

The product

$(n)$ of any

$5$ consecutive numbers will always be divisible by

$120$ and its factors.

$n(n+1)(n+2)(n+3)(n+4)$

$5\times 4\times 3\times 2=120$

Answer C:

$f(x)=1-x+x^2-x^3+....-x^{15}+x^{16}-x^{17}$ , then the coefficient of

$x^2$ in

$f(x-1)$ is?

0%
$826$
0%
$816$
0%
$822$
0%
None of these

Explanation

$f\left( x \right) = 1 - \left( {x - 1} \right) + {\left( {x - 1} \right)^2} - {\left( {x - 1} \right)^3} + {\left( {x - 1} \right)^4}.... - {\left( {x - 1} \right)^{15}} + {\left( {x - 1} \right)^{16}} + {\left( {x - 1} \right)^{17}}$

The coefficient of

${x^2}$ by using binomial expansion

$^2{c_0}{x^2}{ + ^3}{c_1}{x^2}{ + ^4}{c_2}{x^2}{ + ^5}{c_3}{x^2}{ + ^6}{c_4}{x^2}{ + ^7}{c_5}{x^2}{ + ^8}{c_6}{x^2}{ + ^9}{c_7}{x^2}$

${ + ^{10}}{c_8}{x^2}{ - ^{11}}{c_9}{x^2}{ + ^{12}}{c_{10}}{x^2}{ - ^{13}}{c_{11}}{x^2}{ + ^{14}}{c_{12}}{x^2}{ - ^{15}}{c_{13}}{x^2}{ + ^{16}}{c_{14}}{x^2}{ - ^{17}}{c_{15}}{x^2}$

$\Rightarrow {x^2}\left( {1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78 + 91 + 105 + 120 + 136} \right)$

= 372 + 444 = 816

The value of

${ _{ }^{ 13 }{ C } }_{ 2 }+{ _{ }^{ 13 }{ C } }_{ 3 }+{ _{ }^{ 13 }{ C } }_{ 4 }+....+{ _{ }^{ 13 }{ C } }_{ 13 }$ is

0%
${2}^{13}-13$
0%
${2}^{13}-14$
0%
an odd number $\ne {2}^{13}-12$
0%
an even number $\ne {2}^{13}-14$

Explanation

${\textbf{Step -1: Simplifying the problem.}}$

${}^{13}{C_2} + {}^{13}{C_3} + ...... + {}^{13}{C_{13}}.$

$= {}^{13}{C_0} + {}^{13}{C_1} + {}^{13}{C_2} + {}^{13}{C_3} + ...... + {}^{13}{C_{13}} - {}^{13}{C_0} - {}^{13}{C_1}.$

$= \left( {{}^{13}{C_0} + {}^{13}{C_1} + {}^{13}{C_2} + ..... + {}^{13}{C_{13}}} \right) - \left( {{}^{13}{C_0} + {}^{13}{C_1}} \right){\text{ }} \to \left( {\text{i}} \right).$

${\textbf{Step -2: Deduce the value of }}\mathbf{\left( {{}^{13}{C_0} + {}^{13}{C_1} + {}^{13}{C_2} + ..... + {}^{13}{C_{13}}} \right)}{\textbf{ using summation formula}}{\textbf{. }}$

$\sum\limits_{i = 0}^n {{}^n{C_i}} = {2^n}{\text{ }} \to \left( {{\text{ii}}} \right)$

$\Rightarrow \left( {{}^{13}{C_0} + {}^{13}{C_1} + {}^{13}{C_2} + ..... + {}^{13}{C_{13}}} \right) = \sum\limits_{i = 0}^{13} {{}^{13}{C_i}.}$

$\Rightarrow \sum\limits_{i = 0}^{13} {{}^{13}{C_i} = {2^{13}}({\text{By using (}}ii{\text{)}})} .$

${\text{then, using (i) and (ii) we get}},$

$\therefore {2^{13}} - \left( {{}^{13}{C_0} + {}^{13}{C_1}} \right).$

$= {2^{13}} - \left( {\dfrac{{13!}}{{13!0!}} + \dfrac{{13!}}{{12!1!}}} \right).$

$= {2^{13}} - \left( {1 + 13} \right).$

$= {2^{13}} - 14.$

${\textbf{Hence , the correct answer is (B) }}\mathbf{{2^{13}} - 14.}$

A person tries to form as many different parties as he can, out of his

$20$ friends. Each party should consist of the same number. How many friends should be invited at a time? In how many of these parties would the same friends be found?

0%
92378
0%
92364
0%
92376
0%
92391

Explanation

Let the person invite r number of friends at a time. Then, the number of parties is

$^{20}C_r$ , which is maximum when

$r=10$ . If a particular friend will be found in x parties, then x is the number of combinations out of

$20$ in which this particular friend must be included. Therefore, we have to select

$9$ more from

$19$ remaining friends. Hence,

$x=$

$^{19}C_9$ =92378.

A question paper on mathematics consists of twelve questions divided into three parts A, B and C, each containing four questions. In how many ways can an examinee answer five questions, selecting atleast one from each part.

0%
$624$
0%
$208$
0%
$2304$
0%
None

Explanation

Total no. of ways

$A$

$B$

$C$

$A$

$B$

$C$

$=(^4C_1\times \ ^4C_1 \times \ ^4C_3)\times 3+ (^4C_1\, ^4C_2\, ^4C_2)\times 3$ (

$\therefore$ multiple by 3 as for each A,B,C different combination)

$=192+432$

$=624$

$\frac{{({C_0} + {C_1})({C_1} + {C_2})({C_2} + {C_3}).....({C_{n - 1}} + {C_n})}}{{{C_0} + {C_1}({C_2}......{C_n})}} =$

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

In an examination there are three multiple choice questions and each question has

$4$ choices. Number of ways in which a student can fail to get all answer correct is?

0%
$11$
0%
$12$
0%
$27$
0%
$63$

Explanation

There are 4 choices for each question. So, there

are 4 ways to answer a question.

Number of ways to answer 3 questions

$=4\times 4\times 4$

$\downarrow$

$1^{st}$ question

$2^{nd}$

$3^{rd}$

$= 64$

Out of 64 ways there is only one way which has

all the answers correct.

So, Number of ways in which a student can fail to get all answers

correct

$= 64-1$

$= 63$

$7$ boys and

$8$ girls have to sit in a row on

$15$ chairs numbered from

$1$ to

$15$ then?

0%
Number of ways boys and girls sit alternately is $8!7!$
0%
Number of ways boys and girls sit alternately is $2(8!7!)$
0%
The number of ways in which first and fifteenth chair are occupied by boys and between any two boys an even number of girls sit is $^9C_4$ $8!7!$
0%
The number of ways in which first and last seat are occupied by boys and between any two boys an even number of girls sit is $(2{^9C_4}8!7!)$ .

Explanation

$\begin{matrix} G-denotes\, \, Girls \\ X-denotes\, \, boys \\ \Rightarrow G\times G\times G\times G\times G\times G\times G\times G \\ \Rightarrow Number\, \, of\, \, ways\, \, of\, Girls\, \, is\, \, 8!\, \, alternate \\ \Rightarrow Number\, \, of\, \, ways\, \, of\, boys\, \, is\, \, 7! \\ \, \, \, \, \, \, \, after\, \, late \\ \Rightarrow Number\, \, of\, \, boys\, \, and\, \, girls\, \, sit\, \, alternate\, \, =8!7! \\ When\, \, first\, \, and\, \, fifteen\, \, chair\, \, are\, \, fixed\, \, and\, \\ \, first\, \, chair\, \, \times G\times G\times G\times G\times G\times G\times G\times fifteen\, \, chair\, \, between\, \, any\, \, two\, \, boys \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \underline { 9{ C_{ 4 } }8!7! } \, \, \, Ans. \\ \end{matrix}$

Number of different paths of shortest distance from A to B in the grid which do not pass through M.

0%
$462$
0%
$262$
0%
$442$
0%
$^{11}C_5-[^6C_3\times^5C_2]$

Explanation

$\displaystyle \dfrac { 6! }{ 3!3! } =\dfrac { 6.5.4 }{ 3.2.1 } =20$

$\displaystyle \dfrac {11!}{6! 5!} = \dfrac { 11.10.9.8.7. }{ 5.4.3.2.1 }$

$\displaystyle 11\times42 = 462$

State following are True or False

If m=n=p and the groups have identical qualitative characterstic then the number of groups

$=\dfrac { (3n)! }{ n!n!n!3! }$
Note : If 3n different things are to be distributed equally three people then the number of ways

$=\dfrac { (3n)! }{ { (n!) }^{ 3 } }$

0%
True
0%
False

Explanation

If the groups are identical there is no chance of arrangement so number of groups is

$\dfrac{(3{n}!)}{n!{n!}{n!}3!}$

If the groups are not identical then groups can also be arranged so number of groups is

$\dfrac{(3{n}!)}{{n!}{n!}{n!}}$

Hence it is

$True$

A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing the element of P.A subset Q of A is again chosen. The number of way of choosing P and Q so that P Q =

$\phi$ is :-

0%
$2^{2n}-^{2n}C_{n}$
0%
$2^{n}$
0%
$2^{n}-1$
0%
$3^{n}$

Explanation

Let A =

${a_{1},a_{2},a_{3},....a_{n}}$ . For

$a_{1}$

$\epsilon$ For

$a_{1} \epsilon$ A we have following choices:
(i)

$a_{1} \epsilon$ P and

$a_{1} \epsilon$ Q
(ii)

$a_{1} \epsilon$ P

$\notin$
(iii)

$_{1}\notin P and a_{1}\epsilon Q$
(Iv)

$_{1}\notin P and a_{1}\notin Q$
Out of these only (Ii), and (iii) and (iv) imply

$a_{1}\notin P\cap Q$ therefore, the number of ways in which none of

$a_{1},a_{2}.....a_{n}$ belong

$P\cap Q$ is

$3^{n}$ .

A box contains 5 pairs of shoes. If 4 shoes are selected, then the number of ways in which exactly one pair of shoes obtained is :

0%
120
0%
140
0%
160
0%
180

Explanation

We have,

A box contains pair of shoes $=5$

Selected pair of shoes $=4$

Then,

Exactly one pair of shoes obtained is

${{=}^{5}}{{P}_{4}}$

$=\dfrac{5!}{\left( 5-4 \right)!}$

$=\dfrac{5!}{1!}=5!$

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

$=120$

Hence, this is the answer.

Let

$5 < n_1 < n_2 < n_3 < n_4$ be integers such that

$n_1+n_2+n_3+n_4=35$ . The number of such distinct arrangements

$(n_1, n_2, n_3, n_4)$ .

0%
$^{38}C_3$
0%
$^8C_3$
0%
$5$
0%
$6$

Explanation

${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+.....+{ n }_{ k }=R$

for this arrangement,

No. of arrangement or distinct arrangements are

${ R+K-1 }_{ { C }_{ K-1 } }$

So, for

${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+{ n }_{ 4 }=35$

No. of arrangements

$={ 35+4-1 }_{ { C }_{ 4-1 } }={ 38 }_{ { C }_{ 3 } }$

Let

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

$C_r=$

$^{n}C_r$ ). On the basis of information, answer the following question.

$2(C_2)-4(C_4)+6(C_6)$ _______ is?

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

Given

$4$ flags of different colours, how many different signals can be generated. If a signal requires the use of

$2$ flags one below the other?

0%
$4$
0%
$3$
0%
$12$
0%
$1$

Explanation

No. of ways of selecting two flags out of four =

$^4C_2$

So, total possible different signals generated =

$^4C_2\times2!$

$\implies 6\times2=12$

There are

$k$ different books and

$l$ copies of each in a college library. The number of ways in which a student can make a selection of one or more books is

0%
$(k+1)^{l}$
0%
$(l+1)^{k}$
0%
$(k+1)^{l}-1$
0%
$(l+1)^{k}-1$

There are

$4$ letter boxes in a post office. In how many ways can a man post

$8$ distinct letters?

0%
$4\times 8$
0%
${8}^{4}$
0%
${4}^{8}$
0%
$P\left (8,4\right)$

Explanation

$\rightarrow$ Each letter can go to 4 boxes

$\therefore$ Total number of ways

$= 4\times 8 = 32$

1119315_1167002_ans_77c0fe408ea84306b35b2b613f0796ac.jpeg

$4$ buses runs between Bhopal and Gwalior. If a man goes from Gwalior to Bhopal by a bus and comes back to Gwalior by another bus, then the total possible ways are

0%
$12$
0%
$16$
0%
$4$
0%
$8$

Explanation

$\rightarrow$ On going Bhopal to Gwalior

he has 4 options

On returning he has

(4-1) options = 3

$\therefore$ Total ways

$= 4 \times3$ = 12 ways

Number of five-digit numbers divisible by 5 that can be formed from the digits

$0,1, 2, 3, 4, 5$ without repetition of digits are

0%
$240$
0%
$360$
0%
$148$
0%
$216$

Ten persons, amongst whom are

$A$ ,

$B$ and

$C$ to speak at a function. The number of ways in which it can be done if

$A$ wants to speak before

$B$ AND

$B$ wants to speak before

$C$ is

0%
$\dfrac{{10!}}{6}$
0%
$3!$ $7!$
0%
$^{10}{P_3}.7!$
0%
none of these

There are n locks and n matching keys. If all the locks and keys are to be perfectly matched, find the maximum number of trails required to open a lock.

0%
$^{n}C_{2}$
0%
$\sum _{ k=2 }^{ n }{ \left( k+2 \right) }$
0%
$\dfrac{n(n+1)}{2}$
0%
$^{n+1}C_{2}$

The number of ways in which a mixed doubles tennis game can be arranged between 10 players consisting of 6 men and 4 women is .

0%
180
0%
90
0%
48
0%
12

Explanation

Number of ways to select two men out of six men

$= ^{6}C_{2}$

Number of ways to select two women out of four

women

$= ^{4}C_{2}$

Number of ways of selecting the players for the

mixed double

$= ^{6}C_{2}\times ^{4}C_{2} = 15\times 16 = 90$

Let

$M_{1},M_{2},W_{1}\& W_{2}$ are selected players for we

mixed double tennis game

$M_{1}$ chooses

$W_{1}$ then

$M_{2}$ has

$W_{2}$ as the partner or if

$M_{1}$ chooses

$W_{2}$ , lean

$M_{2}$ has

$W_{1}$ as the partner

$\therefore$ There are

$2$ choice for the teams.

Thus, Number of ways in which mixed double

tennis game be arranged

$90\times 2 = 180$

1479265_1182649_ans_5c80a1327f534b2cb3cafee6a605f578.jpg

Let

$p=\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{5}{1\times6}+.......+\dfrac{1}{2013\times2014}$ and

$Q=\dfrac{1}{1008\times2014}+\dfrac{1}{1009\times2013}+.........+\dfrac{1}{2014\times1008}$
then

$\dfrac{P}{Q}=$

0%
$2013$
0%
$2014$
0%
$1511$
0%
$2$

The number of intersection points of diagonals of

$2009$ sides polygon, which lie inside the polygon.

0%
$^{2009}C_4$
0%
$^{2009}C_2$
0%
$^{2008}C_4$
0%
$^{2008}C_2$

Explanation

Choosing any

$4$ vertices make

$1$ intersection

$\therefore \$ No. of intersection points of diagonals

$=^nC_4$

Here

$n=2009$ , No. of points

$=^{2009}C_4$

$\therefore \$ Option

$A$ is correct.

1371791_1169558_ans_29bde493772948c8bb3039f9f24e7f76.png

Let the eleven letters,

$A, B, ....K$ denote an artbitrary permutation of the integers

$(1,2,....11)$ , then

$(A-1)(B-2)(C-3)...(K-11)$ is

0%
Necessarily zero
0%
Always odd
0%
Always evem
0%
None of these

$1+2+3+......t_n= ^{n+1}P_2$

0%
True
0%
False

Consider all permutations of the letters of the word MORADABAD.

The number of permutations which contain the word BAD is:

0%
$21 \times 5!$
0%
$7 \times 5!$
0%
$6 \times 5!$
0%
$2 \times 5!$

Explanation

Total number of letters =

$9$

Consider BAD as a group, then the number of total letters =

$6+1=7$

$\therefore$ Permutation of 7 things =

$7!$

but letters remaining after removing BAD will have two A's

$\therefore$ Total permutation =

$\dfrac{7!}{2}$ =

$21\cdot 5!$

The number of permutation of the letters of the word

$HINDUSTAN$ such that neither the pattern

$'HIN'$ nor

$'DUS'$ nor

$'TAN'$ appears, are :

0%
$166674$
0%
$169194$
0%
$166680$
0%
$181434$

Explanation

Total number of letters

$=9$ in which N is repeated twice

Therefore total number of permutation

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

The number of permutation in which HIN comes as block

$=9-3+1=7!$

The number of permutation in which DUS comes as block

$=7!$

The number of permutation in which TAN comes as block

$=\dfrac{7!}{2}$

The number of permutation in which HIN and DUS comes as block

$=5!$

The number of permutation in which DUS and TAN comes as block

$=5!$

The number of permutation in which HIN and TAN comes as block

$=5!$

Therefore the required number of permutations

$=\dfrac{9!}{2}-\left(7!+7!+\dfrac{7!}{2}-3\times5!+3!\right)$

$=\dfrac{362880}{2}-\left(5040+5040+\dfrac{5040}{2}-360+6\right)$

$=181440-\left(10080+2520-360+6\right)$

$=181440-12246$

$=169194$

In how many ways atleast one horse and atleast one dog can be selected out of eight horses and seven dogs.

0%
${2}^{15}-2$
0%
${2}^{15}-1$
0%
$({2}^{8}-1)({2}^{7}-1)$
0%
${ _{ }^{ 15 }{ C } }_{ 2 }$

Explanation

As we can either select or not select any horse; total ways for horses

$=2^r$ ways to be excluded are when no horse is selected, which is only one way, therefore, favourable ways for horses

$=2^8-1$

similarly, favourable ways for dogs

$=2^7-1$

$\Rightarrow$ Total ways

$=(2^8-1)(2^7-1)$ .

1114491_1196636_ans_42de26e913f0483b830ae0594b35480d.jpg

Determine

$n$ , if

$^ { 2 n } \mathrm { C } _ { 3 } : ^ { n } \mathrm { C } _ { 3 } = 12 : 1$

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

Explanation

$^{2n}C_3 : ^nC_3= 12:1$

first we calculate

$^{2n}C_3$ and

$^nC_3$ seperately

$^{2n}C_3 =\dfrac {2n!}{3!(2n-3)!}=\dfrac {2n(2n-1)(2n-2)(2n-3)!}{3\times 2 \times 1 \times (2n-3)!}$

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

$^nC_3 =\dfrac {n!}{3!(n-3)!}= \dfrac {n(n-1)(n-2)(n-3)!}{3\times 2 \times 1 \times (n-3)!}$

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

$^{2n}C_3 : ^nC_3=12:1$

$\therefore \dfrac {2n(2n-1)(2n-2)}{6}:\dfrac {n(n-1)(n-2)}{6}= 12:1$

$\Rightarrow \dfrac {2n(2n-1)(2n-2)}{n(n-1)(n-2)}=12$

$\Rightarrow 2\times 2 \dfrac {(2n-1)(n-1)}{(n-1)(n-2)}=12$

$\Rightarrow 8n - 4 = 12n - 24$

$\Rightarrow 4n = 20$

$\Rightarrow n=\dfrac {20}{4}$

$\Rightarrow n = 5$

The total number of ways of arranging the letters

$AAABBBCCDEF$ in a row such that letters

$C$ are separated from one another is

0%
$277200$
0%
$138600$
0%
$453600$
0%
none of these

$^{404}C_{4}-^{4}C_{1}^{303}C_{4}+^{4}C_{2}^{202}C_{4}-^{404}C_{4}^{101}C_{4}$ is equal to

0%
$(401)^{4}$
0%
$(101)^{4}$
0%
$0$
0%
$(201)^{4}$

The number of ways in which the letters of the word

$"ARRANGE"$ can be permuted such that

$R's$ occur together is

0%
$\dfrac{!7}{!2!2}$
0%
$6!$
0%
$\dfrac{6!}{2!}$
0%
none of these

The number of positive integral solutions of the equation

$x _ { 1 } x _ { 2 } x _ { 3 } x _ { 4 } x _ { 5 } = 1050$ is

0%
1800
0%
1600
0%
1400
0%
1875

Explanation

We have,

${x_1}{x_2}{x_3}{x_4}{x_5} = 1050 = 2 \times 3 \times {5^2} \times 7$

Now,

$2,3$ and

$7$ can be put in any boxes

${x_1},{x_2},{x_3},{x_4}$ and

${x_5}$ .

Also

$5,5$ can be distributed in

$5$ boxes in

$^{2 + 5 - 1}{C_{5 - 1}}{ = ^6}{C_4} = 15$ Ways.

So total no. of positive integral solutions

$=15 \times5 \times5 \times5$

$=1875$

Option

$D$ is correct answer.

$(1+x+x^2)^n=\displaystyle\sum^{2n}_{r=0}a_rx^r$ , then

$a_0a_{2r}-a_1a_{2r+1}+a_2a_{2r+2}-....=?$

0%
$a_r$
0%
$a_{n-2r}$
0%
$a_{n+r}$
0%
$a_{2r}$

If repetitions are not allowed, the number of numbers consisting of

$4$ digits and divisible by

$5$ and formed out of

$0,1,2,3,4,5,6$ is

0%
$220$
0%
$240$
0%
$370$
0%
$588$

Explanation

$\rightarrow$ For divisibility by

$5$ , unit's place should be

$0$ or

$5$

with

$0$ , ----

$0$

$6$ choices

$\Rightarrow ^6C_3\times 3!$

$=\dfrac{6!}{3!3!}\times 3!=6.5.4$

$=120$

with

$5$ , ----

$5$

$6$ choices

$=120$

$^5C_2\times 2!=10\times 2=20$

$\Rightarrow 120+120-20=220$

$\rightarrow (A)$

Number of cyphers at the end of 202

$\mathrm { C } _ { 1001 }$ is ________.

0%
0
0%
1
0%
2
0%
None

Letters of the word

$MATHEMATICS$ are arranged in all the possible ways, in how many words letter

$C$ is between

$S$ and

$H$ (these three letter are not necessary together)?

0%
$(11!)/(2!)(2!)(2!)$
0%
$(11!)/(3!)(2!)(2!)$
0%
$(11!)/(3!)(3!)(2!)$ `
0%
$None\ of\ these$

${ S }_{ n }={ C }_{ 0 }{ C }_{ 1 }+{ C }_{ 1 }{ C }_{ 2 }+.......+{ C }_{ n-1 }{ C }_{ n }$ and

$\frac { { S }_{ n+1 } }{ { S }_{ n } } =\frac { 15 }{ 4 }$ , then value of n

0%
3, 7
0%
2, 4
0%
1, 3
0%
1, 2

$x+y=1$ , then

$\displaystyle\sum^n_{r=0}r\cdot {^{n}C_r}x^r\cdot y^{n-r}=?$

0%
$1$
0%
n
0%
nx
0%
ny

$\displaystyle\sum^{n-r}_{k=1}$

$^{n-k}C_r={^{x}C_y}$ then?

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

The number of permutations of letters of the word "PARALLAL" atken four at a time must be,

0%
$216$
0%
$244$
0%
$286$
0%
$1680$

Explanation

Permutation of word PARALLAL taken four at a time

Four letter word might be

$LLL$ _

So total of

$16$ words [As

$4$ ways,

$4$ spots]

For

$AALL$ ,

$6$ different arrangements

$AA$ _ _, can be

${ 4 }_{ { C }_{ 2 } }$ ways

$=6$ ways

Arranged in

$12$ ways is total of

$72$ words

And extra words can be formed in

$120$ ways

$16+6+144+120=286$ ways

The value of

$\sum _{ r=0 }^{ s }{ \sum _{ s=1 }^{ n }{ } } ^nC_s.$

$^sC_r$ (where

$r\le s$ is )

0%
$3^n$
0%
$n.3^{n-1}$
0%
$n.3^{n-1}-1$
0%
$3^n-1$

$\displaystyle \frac{1}{{^4{C_n}}} = \frac{1}{{^5{C_n}}} + \frac{1}{{^6{C_n}}}$ , then

$n=$

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

Explanation

$\dfrac { 1 }{ { 4 }_{ { C }_{ n } } } =\dfrac { 1 }{ { 5 }_{ { C }_{ n } } } +\dfrac { 1 }{ { 6 }_{ { C }_{ n } } }$

$\Rightarrow \dfrac { 1 }{ \dfrac { 4! }{ \left( 4-n \right) !n! } } =\dfrac { 1 }{ \dfrac { 5! }{ \left( 5-n \right) !n! } } +\dfrac { 1 }{ \dfrac { 6! }{ \left( 6-n \right) !n! } }$

$\Rightarrow \dfrac { \left( 4-n \right) ! }{ 4! } =\dfrac { \left( 5-n \right) ! }{ 5! } +\dfrac { \left( 6-n \right) ! }{ 6! }$

$\Rightarrow \dfrac { 1 }{ 4! } =\dfrac { 5-n }{ 5! } +\dfrac { \left( 6-n \right) \left( 5-n \right) }{ 6! }$

$\Rightarrow \dfrac { 5-n }{ 5 } +\dfrac { \left( 6-n \right) \left( 5-n \right) }{ 6\times 5 } =1$

$\Rightarrow \left( 5-n \right) \left( \dfrac { 1 }{ 5 } +\dfrac { 6-n }{ 30 } \right) =1$

$\Rightarrow \left( 5-n \right) \left[ \dfrac { 12-n }{ 30 } \right] =1$

$\Rightarrow \left( 5-n \right) \left( 12-n \right) =30$

$\Rightarrow { n }^{ 2 }-17n+30=0$

$\Rightarrow n=15$ or

$n=2$

$\because$

$n$ cannot be greater than

$4$

$\therefore$

$n=2$ [B]

There are n identical red balls & m identical green balls. The number of different linear arrangements consisting of "n red balls but not necessarily all the green balls" is

${^{x}C_y}$ then?

0%
$x=m+n, y=m$
0%
$x=m+n+1, y=m$
0%
$x=m+n+1, y=m+1$
0%
$x=m+n. y=n$

The exponent of 11 in

$^{200}C_{125}$ is

0%
2
0%
1
0%
4
0%
5

The number of ways of 3 scholarship of unequal value be awarded to 17 candidates, Such that no candidate gets more than one scholarship is

0%
$^{ 17 }{ C }_{ 3 }$
0%
${ 17 }^{ 3 }$
0%
${ 3 }^{ 17 }$
0%
$^{ 17 }P_{ 3 }$

Explanation

There are

$3$ scholarships to be awarded between

$17$ candidates so that no one candidate can get more than one scholarship.

The

$1st$ scholarship can be given to any of the

$17$ candidates, So

$17$ choices.

$\Rightarrow$ The

$2nd$ can be given to the remaining

$16$ candidates in

${^{16}C_1}=16$ ways.

Similarly the third in

$15$ ways.

$\Rightarrow Total =17\times16\times15={^{17}P_3}$

Hence, the answer is

${^{17}P_3}$

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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