MCQ Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 2

From a well shuffled pack of

$52$ playing cards two cards drawn at random. The probability that either both are red or both are kings is:

0%
$\dfrac{{\left( {^{26}{C_2} + \,{\,^4}{C_2}} \right)}}{{^{52}{C_2}}}$
0%
$\dfrac{{\left( {^{26}{C_2} + {\,^4}{C_2} - {\,^2}{C_2}} \right)}}{{^{52}{C_2}}}$
0%
$\dfrac{{^{30}{C_2}}}{{^{52}{C_2}}}$
0%
$\dfrac{{^{39}{C_2}}}{{^{52}{C_2}}}$

Explanation

Assuming, cards are drawn without replacement:

Total possible events $=^{52}C_2$

P(both red) $=\dfrac{ ^{26}C_2}{^{52}C_2}$

P(both king) $=\dfrac{^{4}C_2}{^{52}C_2}$

P(both red & king) $=\dfrac {^2C_2}{{^{52}C_2}}$

We use the basic addition rule,

P(both black or both queens) = P(both red) + P(both queens) - P(both black as well as queens)

Required probability $=\dfrac{^{26}C_2+^{4}C_2-^2C_2}{^{52}C_2}$

Number of odd numbers of five distinct digits can be formed by the digits

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

0%
$24$
0%
$120$
0%
$48$
0%
$36$

Explanation

We have total digits $0,1,2,3,4.$

We want to form odd number, then at last digit can become $1$ or $3$ .

So, two number are fixed for last.

Now, we have option for $1st, 2nd, 3rd$ and $4th$ place are $3, 3, 2, 1$

So , number of getting odd numbers using given five distinct digit $=3\times 3\times 2\times 1\times 2=36$

Hence, this is the answer.

The number of ways in which ten candidates

$A_1, A_2,......A_{10}$ can be ranked such that

$A_1$ is always above

$A_{10}$ is

0%
$5!$
0%
$2(5!)$
0%
$10!$
0%
$\dfrac{1}{2}(10!)$

Explanation

Total number of ways to arrange

$A_1,A_2 .. A_{10} = 10!$

$\dfrac{10! }{ 2}$ combinations

$A_1$ will be above

$A_2$ and in the other half

$A_2$ will be above

$A_1$ .

Ans :

$\dfrac{10! }{2}$

$3$ letters are posted in

$5$ letters boxes. If all the letters are not posted in the same box, then number of ways of posting is

0%
$120$
0%
$125$
0%
$130$
0%
$124$

Explanation

According to problem,

$3$ letters are posted

$5$ boxes

implies that,

we have a letter and can be posted in any of the

$5$ boxes.

Similarly next letter can be posted in 5 ways, and the all other follow same method

implies that

${\left(5\right)}^3$

$= 5 \times 5 \times 5$

Hence,

$125$ possible ways for posting in

$5$ boxes.

There are

$10$ trees between two stations

$A$ and

$B$ . Three of them are to be cut down then the total number of ways so that no two trees are to be cut consecutively, is

0%
$^{8}C_{3}$
0%
$^{7}C_{3}$
0%
$^{10}C_{3}$
0%
$^{9}C_{3}$

Explanation

$3$ trees to be cut

$7$ left

Total no. of ways to cut

$3$ trees

$=$

Total no. of was to place

$3$ trees

in the given gaps,

$=$ select

$3$ gaps from

$8$

$=\, ^8C_3$

1346343_1237523_ans_da4b52913ed64b1c8f7d9fb78003be92.png

A _____ is an arrangement of all or part of set of object in a definite order.

0%
permutation
0%
function
0%
combination
0%
factorial

$^{n + 1}{C_3} = 4\,{\,^n}{C_2}$ then

$n=$

0%
$12$
0%
$10$
0%
$16$
0%
$11$

Explanation

Given that, $^{n+1}{{C}_{3}}={{4.}^{n}}{{C}_{2}}$

Then $n=?$

$^{n+1}{{C}_{3}}={{4.}^{n}}{{C}_{2}}$

$\Rightarrow \dfrac{\left( n+1 \right)!}{3!\left( n+1-3 \right)!}=4.\dfrac{n!}{2!\left( n-2 \right)!}$

$\Rightarrow \dfrac{\left( n+1 \right)n!}{3\times 2!\left( n-2 \right)!}=4.\dfrac{n!}{2!\left( n-2 \right)!}$

$\Rightarrow \dfrac{\left( n+1 \right)}{3}=4$

$\Rightarrow n+1=12$

$\Rightarrow n=12-1=11\,$

Hence, this is the answer.

Option (D) is correct.

If the letter of the word

$LATE$ be permuted and the words so formed be arranged as in a dictionary . Then the rank of

$LATE$ is :

0%
$12$
0%
$13$
0%
$14$
0%
$15$

Explanation

LATE

No. of words that start with 'A'

$\Rightarrow 3!=6$

No. of words that start with E

$\Rightarrow 3!=6$

$6+6=12$

$13^{th}$ word starts with L

$13^{th}$ word = LAET

$14^{th}$ word = LATE

$\therefore$ Rank of LATE is

$14$

$\left( {n + 1} \right)! = 12 \times (n - 1)!\;then\;n =$

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

Explanation

We have,

$\left( {n + 1} \right)! = 12 \times (n - 1)!$

$(n + 1)\times n\times (n-1)! = 12 \times (n - 1)!$

$(n + 1)\times n = 12$

$n^2 +n- 12=0$

$n^2 +4n-3n- 12=0$

$(n+4)(n-3)=0$

$(n+4)\ne 0$

So,

$n-3=0$

$n=3$

Hence, this is the answer.

How many chords can be drawn through 21 points on a circle ?

0%
$301$
0%
$210$
0%
$111$
0%
$220$

Explanation

A chord is obtained by joining any two points on a circle. Therefore, total number of chords drawn through 21 points is same as the number of ways of selecting 2 points out of 21 points. This can be done in

$21_{C_2}$ ways.

Hence, total number of chords =

$^{21}{C_2}=\dfrac{21!}{19!2!}=21 \times 10 = 210$

${2n_{C}}_{4} : {n_{C}}_{3} = 21:1$ , then find the value of n.

0%
4
0%
5
0%
6
0%
7

Explanation

$\dfrac{{2n_{C}}_{4}}{{n_{C}}_{3}}=\dfrac{21}{1}$

${2n_{C}}_{4}=21{n_{C}}_{3}$

$\dfrac{(2n)!}{(2n-4)!4!}=21\dfrac{n!}{(n-3)!3!}$

$2n(2n-1)(2n-2)(2n-3)=84n(n-1)(n-2)$

$4(2n-1)(2n-3)=84(n-2)$

From the given options,

$n=5$ satisfies the above equation.

Hence, option B is correct.

$^{ 28 }{ C }_{ 2r }:^{ 24 }{ C }_{ 2r }=225:11$ , then find the value of

$r$

0%
$r = 4$
0%
$r = 3$
0%
$r = 7$
0%
$r = 8$

If P (n, n) denotes the number of permutations of n different things taken all at a time then P (n, n ) is also identical to:

0%
$P ( n - 1 , n - 1 )$
0%
$P ( n , n - 1 )$
0%
$\mathrm { r } ! \mathrm { P } ( \mathrm { n } , \mathrm { n } - \mathrm { r } )$
0%
$( n - r ) \cdot P ( n , r )$

The no .of ways of selecting

$3$ men and

$2$ women from

$6$ men and

$6$ women.

0%
$^6C_3 ^6C_2$
0%
$^{12} C_5$
0%
$^6C_5$
0%
None of these

Explanation

The no.o f ways of selecting Men is

$^6C_3$

The no.o f ways of selecting Women is

$^6C_2$

Total no. of ways is

$^6C_3\times ^6C_2$

How many six letter words be made out of the letters of

$ASSIST$ ? In how many words the alphabets

$S$ alternates with other letters ?

0%
$120,6$
0%
$720,\,12$
0%
$120,\,12$
0%
$720,\,24$

Explanation

The word ASSIST has 3 's' and 3 other characters (AIT).

Place the other 3 characters first. This can be done in 3! ways.

Now there are 4 spaces remaining (1234) mark as 1,2,3,4.

3 's' 5 can be placed at the spaces marked as 1,2,3 or at the spaces marked as 2,3,4 (2)was

Total no.of ways =3! .2

$=3\times 2\times 2=12$

Total no.of ways =5!.

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

$\therefore$ so, the answer is 120,12.

1219395_1441079_ans_7bbe7e8c89714c6d990173d3ae9c3d88.jpg

The number of ways in which 6 rings can be worn on the four fingers of one hand is

0%
${ 4 }^{ 6 }$
0%
$^{ 6 }{ C }_{ 4 }$
0%
${ 6 }^{ 4 }$
0%
None of these

Explanation

The four fingers can be selected in

$^5C_4=5$ ways

And 6 rings can be placed in

$^6C_4$ ways

So total no.of ways are

$5\times ^6C_4$

If the coefficients of three consecutive terms in the expansion of

$(1+x)^n$ are in the ratio of

$1:7:42$ , then

$n$ is divisible by-

0%
95
0%
55
0%
35
0%
11

Explanation

Let the three consecutive terms be

$(r-1)th, rth , (r+1)th$ terms.

That is,

$T_{r-1}, T_{r}. T_{r+1}$ .

We know that general term of expansion

$(a+b)^{n}$ is

$T_{r+1}={n_{C}}_{r}a^{n-r}b^{r}$

For

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

Putting

$a=1, b=x$ ,

$T_{r+1}={n_{C}}_{r}1^{n-r}x^r$

$T_{r+1}={n_{C}}_{r}x^r$

Therefore, coefficient of

$(r+1)th$ term =

${n_{C}}_{r}$

For

$rth$ term of

$(1+x)^{n}$

Replacing r with r-1, we get,

$T_{r}={n_{C}}_{r-1}x^{r-1}$

Therefore, coefficient of

$rth$ term =

${n_{C}}_{r-1}$

For

$(r-1)th$ term of

$(1+x)^{n}$

Replacing r with r-2. we get,

$T_{r-1}={n_{C}}_{r-2}x^{r-2}$

Therefore, coefficient of

$(r-1)th$ term =

${n_{C}}_{r-2}$

Since the coefficient of (r-1)th, rth and (r+1)th terms are in the ratio 1 : 7 : 42,

$\dfrac{{n_{C}}_{r-2}}{{n_{C}}_{r-1}}=\dfrac{1}{7}$

$\dfrac{\dfrac{n!}{(r-2)![n-(r-2)]!}}{\dfrac{n!}{(r-1)![n-(r-1)]!}}=\dfrac{1}{7}$

$\dfrac{(r-1)[n-(r-1)]!}{[n-(r-2)]!}=\dfrac{1}{7}$

$\dfrac{(r-1)(n-r+1)!}{(n-r+2)!}=\dfrac{1}{7}$

$\dfrac{(r-1)(n-r+1)!}{(n-r+2)(n-r+1)!}=\dfrac{1}{7}$

$\dfrac{(r-1)}{(n-r+2)}=\dfrac{1}{7}$

$n-8r+9=0$ ......(1)

Also,

$\dfrac{\dfrac{n!}{(r-1)![n-(r-1)]!}}{\dfrac{n!}{r!(n-r)!}}=\dfrac{7}{42}$

$\dfrac{n!}{(r-1)!(n-r+1)!} \times \dfrac{r!(n-r)!}{n!}=\dfrac{1}{6}$

$\dfrac{r(n-r)!}{(n-r+1)!}=\dfrac{1}{6}$

$\dfrac{r}{n+1-r}=\dfrac{1}{6}$

$n-7r+1=0$ .....(2)

Solving equations 1 & 2, we get,

$r=8$ and

$n=55$

Hence,

$n=55$

When we realize a specific implementation of a pancake algorithm, every move when we find the greatest of the sized array and flipping can be modeled through ____________.

0%
Combinations
0%
Exponential functions
0%
Logarithmic functions
0%
Permutations

Two persons entered a Railway compartment in which 7 seats were vacant.The number of ways in which they can be seated is

0%
$30$
0%
$42$
0%
$720$
0%
$360$

Explanation

$\rightarrow$ The

$1^{st}$ person can take one of the 7 seats

$\rightarrow$ The

$2^{nd}$ person can take any one of the remaining 6 seats.

$\Rightarrow$ So, the total

$=7\times 6=42$

No. of permutations of

$25$ dissimilar things taken more than

$15$ at a time when repetitions are allowed is

0%
$\dfrac{25}{24}({25}^{25}-{25}^{15})$
0%
$\dfrac{25}{24}({25}^{25}-{25}^{10})$
0%
$\dfrac{25}{24}({25}^{25}+{25}^{15})$
0%
$\dfrac{25}{24}({25}^{25}+{25}^{10})$

Explanation

No of permutations of 16 objects

$= 25^{16}$
No of permutations of 17 objects

$= 25^{17}$
18

$\rightarrow 25^{18}$
19

$\rightarrow 25^{19}$
25

$\rightarrow 25^{25}$
So, S

$=$

$25^{25}+25^{24}+................+25^{16}$

$=$

$\dfrac{25^{16}[25^{10}-1]}{25-1}$

$\Rightarrow$ Ans

$=$

$\dfrac{25}{24}(25^{25}-25^{15})$

There are 8 types of pant pieces and

$9$ types of shirt pieces with a man. The number of ways in which a pair (

$1$ pant,

$1$ shirt) can be stitched by the tailor is

0%
$17$
0%
$56$
0%
$64$
0%
$72$

Explanation

Each piece of pant can be paired with

$9$ different pieces of shirt.
Now there are

$8$ pieces of pant.
Taking one piece of a pant at time, there are

$9$ ways to stitch it, since there are

$9$ different pieces of shirt.
Hence, the total ways will be

$=8\times 9$

$=72$

$^nP_r = ^nP{_r}{_+}{_1}$ and

$^nC_r = ^nC{_r}{_-}{_1}$ , then the values of n and r are:

0%
$r , 3$
0%
$3, 2$
0%
$4, 2$
0%
$3, 4$

Explanation

$^nP_r = ^nP_{r+1}$

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

$(n-r-1)! = (n-r)!$

$n-r = 1$ -------------(1)

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

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

$(r-1)!(n+1-r)! = (n-r)!r!$

$n+1-r = r$

$n+1 = 2r$

$n = 2r-1$ -------------(2)

By solving both equations, we get

$r = 2$

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

Using the digits

$0, 2, 4, 6, 8$ not more than once in any number, the number of

$5$ digited numbers that can be formed is

0%
16
0%
24
0%
120
0%
96

Explanation

Total number of 5 digit numbers will be 5!.
Among these the ones starting with 0, are 4!.

Hence the total number of 5 digit numbers will be

$5!-4!$

$=4!(4)$

$=24\times 4$

$=96$

The number of different signals that can be formed by using any number of flags from

$4$ flags of different colours is

0%
$24$
0%
$256$
0%
$64$
0%
$60$

Explanation

Signals of

$1$ flag

$= \ ^4C_1 = 4$

$2$ flags

$=\ ^4C_2 2! = 12$

$3$ flags

$=\ ^4C_33! = 24$

$4$ flags

$=\ ^4C_44! = 24$
Therefore, total

$=64$ .

The product of

$n$ consecutive natural numbers is always divisible by

0%
$4n!$
0%
$3n!$
0%
$2n!$
0%
$n!$

Explanation

Let

$n$ consecutive natural numbers be

$(k+1), (k+2)........(k+n)$ , where

$k\in I >0$
Then product of these numbers are

$=(k+1)(k+2)........(k+n)=P$ (say)

$\Rightarrow P = \dfrac{k!(k+1)(k+2)........(k+n)}{k!}=\dfrac{(k+n)!}{k!}=\dfrac{(k+n)!}{k!n!}(n!) = (n!)\cdot ^{n+k}C_n$
Hence, product of

$n$ natural numbers will be always divisible by

$n!$ .

The number of words that can be formed using any number of letters of the word "KANPUR" without repeating any letter is

0%
$720$
0%
$1956$
0%
$360$
0%
$370$

Explanation

$(a) 1$ Letter word:

Selecting

$1$ letter of

$6$ letter

$K,A,N,P,U,R=\quad^{6}{C}_{1}=6$ ways

$(b) 2$ Letter words:

Selecting

$2$ letter of

$6$ letter

$K,A,N,P,U,R= \quad ^{6}{C}_{2}$

Therefore no. of ways to create

$2$ letter words

$=\quad ^{6}{C}_{2} \times 2!$

$(c) 3$ Letter words:

Selecting

$3$ letter of

$6$ letter

$K,A,N,P,U,R= \quad ^{6}{C}_{3}$

Therefore no. of ways to create

$3$ letter words

$=\quad ^{6}{C}_{3} \times 3!$

$(d) 4$ Letter words:

Selecting

$4$ letter of

$6$ letter

$K,A,N,P,U,R= \quad ^{6}{C}_{4}$

Therefore no. of ways to create

$4$ letter words

$=\quad ^{6}{C}_{4} \times 4!$

$(e) 5$ Letter words:

Selecting

$5$ letter of

$6$ letter

$K,A,N,P,U,R= \quad ^{6}{C}_{5}$

Therefore no. of ways to create

$5$ letter words

$=\quad ^{6}{C}_{5} \times 5!$

$(f) 6$ Letter words:

Selecting

$6$ letter of

$6$ letter

$K,A,N,P,U,R= \quad ^{6}{C}_{6}$

Therefore no. of ways to create

$6$ letter words

$=\quad ^{6}{C}_{6} \times 6!$

Total

$= 6+\quad^{6}{C}_{2}\times 2!+\quad ^{6}{C}_{3}\times 3!+\quad ^{6}{C}_{4}\times 4!+\quad ^{6}{C}_{5}\times 5!+ 6!= 1956$ words

The value of expression

${^4}{^7}C_4 + \sum _{i=1}^{5} {^5}{^{2-i}}C_3$ is:

0%
${^5}{^2}C_4$
0%
${^5}{^2}C_3$
0%
${^5}{^3}C_4$
0%
${^5}{^3}C_3$

Explanation

${^4}{^7}C_4 +$

$\sum \:^{5} _{i=1} \:^{52-i}C_{3}$

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

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

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

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

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

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

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

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

The number of rational numbers

$\dfrac {p}{q}$ , where

$p,q$

$\in$

${1, 2, 3, 4, 5, 6}$ is

0%
23
0%
32
0%
36
0%
63

Explanation

The given 6 digits can occupy the numerator and denominator places in

$\, ^{6}P_{2}=30$ ways. But out of them these sets

$(\displaystyle \frac{1}{2},\frac{2}{4},\frac{3}{6})$ ,

$(\displaystyle \frac{2}{1}$ ,

$\displaystyle \frac{4}{2},\frac{6}{3})$ ,

$(\displaystyle \frac{2}{3}$ ,

$\displaystyle \frac{4}{6})$ ,

$(\displaystyle \frac{3}{2},\frac{6}{4})$ ,

$(\displaystyle \frac{1}{3},\frac{2}{6})$ and

$(\displaystyle \frac{3}{1}$ ,

$\displaystyle \frac{6}{2})$ represent same number. In all above cases we have to consider only one case each.

$\therefore$ no. of rational numbers

$= 30 - (2 + 2+ 1 + 1+ 1 + 1)+1$

$= 22 + 1$ (

$1$ is a rational no)

$= 23$

$\displaystyle ^{14}C_{ 4 }+\sum _{ j=1 }^{ 4 } \quad ^{ (18-j) }C_{ 3 }=$

0%
$^{14}C_5$
0%
$^{18}C_5$
0%
$^{18}C_4$
0%
$^{19}C_4$

Explanation

We know,

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

$^{14}C_{ 4 }+\sum _{ j=1 }^{ 4 } \quad ^{ (18-j) }C_{ 3 }= ^{14}C_{ 4 }+^{17}C_{ 3 }+^{16}C_{ 3 }+^{15}C_{ 3 }+^{14}C_{ 3 }$

$=^{17}C_{ 3 }+^{16}C_{ 3 }+^{15}C_{ 3 }+^{15}C_{ 4 }$

$=^{18}C_{4}$

${^2}{^n}C_3 : ^nC_2 = 44 :3$ , then

$n =$

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

Explanation

Given,

${^2}{^n}C_3 : ^nC_2 = 44:3$

$\Rightarrow \dfrac { 2n! }{ 3!(2n-3)! } : \dfrac { n! }{ 2!(n-2)! } = 44:3$

$\Rightarrow \dfrac {2n! (n-2)!}{3 (2n-3)! n!}= \dfrac{44}{3}$

$\Rightarrow \dfrac { 2n.(2n-1).(2n-2)}{n.(n-1)} =44$

$\Rightarrow 4(2n-1)=4 \times(2 \times 6-1)$

$\Rightarrow n=6$

${^1}{^5}C{_3}{_r} = {^1}{^5}C{_r}{_+}{_3}$ , then

$r=$

0%
$\dfrac{3}{2}$
0%
$3$
0%
$4$
0%
$5$

Explanation

Given,

$^{ 15 }C_{ 3r }=^{ 15 }C_{ r+3 }$

$\Rightarrow 3r+(r+3)=15$

$\Rightarrow 4r=12$

$\Rightarrow r=3$

The number of unsuccessful attempts that can be made by a thief to open a number lock having

$3$ rings in which each rings contains

$6$ numbers is

0%
205
0%
200
0%
210
0%
215

Explanation

Total number of attempts that can be made by thief is equal to placing

$6$ number on

$3$ places (number may be repeated).
Total

$= 6\times 6\times 6 =216$
Hence, total number of unsuccessful attempt

$=216-1 =215$

$n$ 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

$n!\left( 21-n \right) !=21!\dfrac { n!\left( 21-n \right) ! }{ 21! } =\dfrac { 21! }{ { _{ }^{ 21 }{ c } }_{ n }^{ } }$

For minimum value, denominator should be maximum

Maximum value of

${ _{ }^{ 21 }{ c } }_{ n }^{ }={ _{ }^{ 21 }{ c } }_{ \displaystyle\dfrac { 21-1 }{ 2 } }^{ }={ _{ }^{ 21 }{ c } }_{ 10 }^{ }$

So minimum value

$\dfrac { 21! }{ { _{ }^{ 21 }{ c } }_{ 10 }^{ } } =10!11!$

There are 'mn' letters and n post boxes. The number of ways in which these letters can be posted is:

0%
$(mn)^n$
0%
$(mn)^m$
0%
$m{^m}{^n}$
0%
$n{^m}{^n}$

Explanation

Every letter can be posted in any of the n post boxes

$\therefore$ Total number of ways

$=n\times n\times n\times........(mn \ times) = n^{mn}$

The maximum number of persons in a country in which no two persons have an identical set of teeth assuming that there is no person without a tooth is

0%
$2{^3}{^2}$
0%
$2{^3}{^2}$ $- 1$
0%
$32!$
0%
$32! - 1$

Explanation

Total no. of teeth a person has = 32

No. of persons having 1 tooth is =

${ 32 } C_{ 1 }$

No. of persons having 2 tooth is =

${ ^{32}C }_{ 2 }$

No. of persons having 3 tooth is =

${ ^{32}C }_{ 3 }$

...... and so on ..... 32 tooth =

${ ^{32}C }_{ 32 }$

Total persons =

$^{ 32 } C_{ 1 }+ { ^{32}C }_{ 2 } + { ^{32}C }_{ 3 }........ + { ^{32}C }_{ 32 }\\ = { 2 }^{ 32 }-1$

$^nC{_{r-1}}= 36, ^nC_r=84, ^nC{_{r+1}}= 126$ , then

$(n,r)$

$=$

0%
$(9,6)$
0%
$(9,5)$
0%
$(9,3)$
0%
$(9,2)$

Explanation

$\dfrac { ^{ n }C_{ r } }{ ^{ n }C_{ r-1 } } =\dfrac { n-r+1 }{ r } =\dfrac { 84 }{ 36 } =\dfrac { 7 }{ 3 } \Rightarrow 3n-10r+3=0$
and

$\dfrac { ^{ n }C_{ r+1} }{ ^{ n }C_{ r } } =\dfrac { n-r }{ r+1 } =\dfrac { 126 }{ 84 } =\dfrac { 3 }{ 2 } \Rightarrow 2n-5r-3=0$
Solving both equations for

$n$ and

$r$ ,

$(n.r)=(9,3)$

The number of products that can be formed with

$8$ prime numbers is:

0%
$247$
0%
$252$
0%
$5$
0%
$248$

Explanation

A product is defined as multiplication between two or more numbers.

Thus, we need to pick up

$2$ numbers,

$3$ numbers,

$4$ numbers and so on from the

$8$ prime numbers till the product of all 8 of those.

$\therefore$ The number of ways become

$^8C_2 + ^8C_3 + ^8C_4 + ^8C_5 + ^8C_6 + ^8C_7 + ^8C_8 = 2^8 - ^8C_0 - ^8C_1$

$= 2^8 - 1 - 8$

$= 247$

A telegraph post has 5 arms, each arm is capable of four distinct positions including the position of rest. The total number of signals that can be made is:

0%
$625$
0%
$1023$
0%
$1024$
0%
$930$

Explanation

From every post 4 positions can be selected, so total number of signals that can be made

$=4^5$

But in that, there will be one case where every position is in rest

So effective number of signals that can be made

$=4^5-1=1023$

Let

$y$ be an element of the set

$A=\left\{1,2,3,5,6,10,15,30\right\}$ and

$x_1,x_2,x_3$ be integers such that

$x_1x_2x_3=y,$ then the number of positive integral solutions of

$x_1x_2x_3=y$ is

0%
$64$
0%
$27$
0%
$81$
0%
None of these

Explanation

Number of solutions of the given equation is same as the number of solutions of the equation

${ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }{ x }_{ 4 }=30=2\times 3\times 5$

Here

${x }_{ 4 }$ is there because if

${ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }=15$ then

${x }_{ 4 }=2$ and if

${ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }=5$ then

${x }_{ 4 }=6$ etc.

${x }_{ 4 }$ is in fact a dummy variable.

Each of

$2,3$ and

$5$ will be a factor of exactly one of

${ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },{ x }_{ 4 }$ in

$4$ ways.

So required number is

${ 4 }^{ 3 }=64$

$m=^n{C_{2}}$ , then

$^m{C_{2}}$ equals

0%
$^{n+1}C_{4}$
0%
$3.^{ n+1}C_{4}$
0%
$^{n}C_{4}$
0%
$^{n+1}C_{3}$

Explanation

$m = ^nC_2 = \dfrac{n(n-1)}{2}=\dfrac{n^2-n}{2}$

$\therefore ^mC_2 =\dfrac{m(m-1)}{2}=\dfrac{1}{2}\dfrac{n^2-n}{2}\left( \dfrac{n^2-n}{2}-1\right)$

$\quad \quad = \dfrac{n(n-1)(n^2-n-2)}{8} =\dfrac{(n+1)(n)(n-1)(n-2)}{8}$

$\quad \quad = 3\cdot \dfrac{(n+1)!}{4!(n-3)!} =3\cdot ^{n+1}C_4$

Match the following:

$\\ A)^{ n }{ P_{ r } }\quad \quad \quad \quad \quad \quad \quad 1)^{n+1}C_{ r }\\ B)^nC_{ r }\quad \quad \quad \quad \quad \quad 2){ \dfrac { n! }{ (n-r)!{r}! } }\\ C)^nC_{ r }+^{n}C_{ r-1 }\quad \quad \quad 3)^nC_{ r }r!\\ D){ \dfrac { ^nC_{ r } }{ ^n{ C_{ r-1 } } } }\quad \quad \quad \quad \quad \quad 4){ \dfrac { r }{ n-r+1 } }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad 5){ \dfrac { n-r+1 }{ r } }$

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

Explanation

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

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

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

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

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

$=^{n+1}C_r$

$\displaystyle \frac{^nC_r}{^nC_{r-1}}=\frac{n-r+1}{r}$

$^{n}P_{r} =$

$30240$ and

$^{n}C_{r} =$

$252$ , then the ordered pair

$(n , r)$

$=$

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

Explanation

Given,

$^{n}P_{r} = 30240$ and

$^{n}C_{r} = 252$

We know,

$\dfrac { ^{ n }P_{ r } }{ ^{ n }C_{ r } } =r!=\dfrac { 30240 }{ 252 } =120=5!$

$\Rightarrow r=5$

$\Rightarrow (n,r)=(10,5)$

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

$n$ is

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

Explanation

Using

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

$\displaystyle ^{n-1}C_{3}+^{n-1}C_{4}=^{n}C_{4}> ^{n}C_{3}$

$\Rightarrow n> 4+3=7$ , where

$n$ is a natural number
Hence, least possible value of

$n$ is

$8$ .

The value of

$\displaystyle E = \frac { (1+17)(1+\frac { 17 }{ 2 } )(1+\frac { 17 }{ 3 } )......(1+\frac { 17 }{ 19 } ) }{ (1+19)(1+\frac { 19 }{ 2 } )(1+\frac { 19 }{ 3 } ).....(1+\frac { 19 }{ 17 } ) }$ is,

0%
$1$
0%
$^{36}C_{17}$
0%
$\dfrac {2}{19}$
0%
$^{36}C_{18}$

Explanation

$\displaystyle E = \frac { (1+17)(1+\frac { 17 }{ 2 } )(1+\frac { 17 }{3 } )......(1+\frac { 17 }{ 19 } ) }{ (1+19)(1+\frac { 19 }{ 2 } )(1+\frac { 19 }{ 3 } ).....(1+\frac { 19 }{ 17 } ) }$

$\quad =\displaystyle \frac{\displaystyle \frac{18\cdot 19\cdot 20.............36}{1\cdot 2\cdot 3\cdot 4...............19}}{\displaystyle \frac{20\cdot 21\cdot 22.............36}{1\cdot 2\cdot 3\cdot 4...............17}}$

$\ \ \ =\displaystyle \frac{18\cdot 19\cdot 20.............36}{1\cdot 2\cdot 3\cdot 4...............19}\times \frac{1\cdot 2\cdot 3\cdot 4...............17}{20\cdot 21\cdot 22.............36}$

$\quad = \displaystyle \frac{18\cdot 19\cdot 20.............36}{20\cdot 21\cdot 22.............36}\times \frac{1\cdot 2\cdot 3\cdot 4...............17}{1\cdot 2\cdot 3\cdot 4...............19}$

$\displaystyle \ \ \ =\frac{18\cdot 19}{1}\times \frac{1}{18\cdot 19}=1$

The expansion

$^n C_r + 4.^nC_{ r-1 } + 6.^nC_{ r-2 }+4.^nC_{r -3 }+^n{ C_{r-4}}=$

0%
$^{ n+4 }C_{ r }$
0%
$2. ^{ n+4 }C_{ r - 1}$
0%
4. $^nc_r$
0%
11. $^nc_r$

Explanation

$\displaystyle^{n}C_{r}+4.\, ^{n}C_{r-1}+6.\, ^{n}C_{r-2}+4\, ^{n}C_{r-3}+\, ^{n}C_{r-4}$

$\displaystyle = ^{4}C_{0}.^{n}C_{r}+^{4}C_{1}.1^{n}C_{r-1}+^{4}C_{2}.^{n}C_{r-2}+^{4}C_{3}.^{n}C_{r-3}+^{4}C_{4}.^{n}C_{r-4}$

$=^{n+4}C_{r}$

The number of rational numbers lying in the interval

$(2002, 2003)$ all of whose digits after the decimal point are non-zero and are in decreasing order is

0%
$\sum _{ i=1 }^{ 9 } 9P_{ i }$
0%
$\sum _{ i=1 }^{ 10 } 9P_{ i }$
0%
$2^{ 9 }-1$
0%
$2^{ 10 }-1$

Explanation

Non zero and decreasing order - That means we just have to choose any number of digits from 1 - 9 (without repetition), then there is only one way to form the rational number.

$\therefore$ No. of rational numbers in

$^{9}C_{1}+\, ^{9}C_{2}+.......+\, ^{9}C_{9}=2^{9}-1$

$^{ n }{ C_{ r-1 } }+^{ n+1 }C_{ r-1 }+^{ n+2 }C_{ r-1 }+.......+^{ 2n }{ C_{ r-1 } },=^{ 2n+1 }C_{ { r^{ 2 }-132 } }-^{ n }C_{ r }$ , then the value of

$r$

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

Explanation

$^nC_{r-1}+ ^{n+1}C_{r-1}+^{n+2}C_{r-1}+...............+^{2n}C_{r-1} = ^{2n+1}C_{r^2-132}- ^nC_r$

$\Rightarrow ^nC_r+^nC_{r-1}+ ^{n+1}C_{r-1}+^{n+2}C_{r-1}+...............+^{2n}C_{r-1} = ^{2n+1}C_{r^2-132}$

$\Rightarrow ^{2n+1}C_r = ^{2n+1}C_{r^2-132}$

$\Rightarrow r = r^2-132$

$\Rightarrow r(r-1) = 132 = 12\times 11$

$\Rightarrow r = 12$

$^nC_{ 3 }=^{ n }C_{ 9 }$ , then

$^n C_{ 2 }=$

0%
66
0%
132
0%
72
0%
98

Explanation

Given,

$^nC_{ 3 }=^{ n }C_{ 9 }$
Now using,

$^{n}C_{r}=^{n}C_{s}$

$\Rightarrow n-r=s$
We have,

$n-3=9$

$\Rightarrow n=12$

$\therefore ^nC_2= ^{12}C_{2}=66$

$n$ and

$r$ are integers such that

$1\le r \le n$ , then

$n . C (n-1, r-1)$

$=$

0%
$C (n, r)$
0%
$n . C (n, r)$
0%
$r C (n, r)$
0%
$(n - 1) . C (n, r)$

Explanation

$\because C(n,r) = \cfrac{n!}{(n-r)!r!}$

$\therefore n . C (n-1, r-1) = n\cdot \cfrac{(n-1)!}{(n-1-r+1)!(r-1)!}$

$\therefore n . C (n-1, r-1)$

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

$\quad = r\cdot \cfrac{n!}{(n-r)!r!}$ , .........................Multiplying with

$'r'$ on both numerator and denominator &

$n\cdot (n-1)!=n!$

$\quad = r\cdot C(n,r)$

$n$ and

$r$ are positive integers such that

$r < n$ , then

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

0%
$^{2n}C_{2r-1}$
0%
$^{(n +1)}C_r$
0%
$^nC_{r+1}$
0%
$^{(n+1)}C_{r+1}$

Explanation

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

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

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

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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