Permutations and Combinations — Page 3 | JEE Mathematics

Q21

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is:

A 880

B 629

C 630

D 879

Correct Answer

Option D

Solution

For alike n objects, number of ways we can select zero or more objects = n + 1 and number of ways we can select at least one object = n Given 10 identical white balls, 9 identical green balls and 7 identical black balls.

To find number of ways for selecting atleast one ball.

Number of ways to choose zero or more white balls = (10 + 1) [since, all white balls are mutually identical] Number of ways to choose zero or more green balls = (9 + 1) [since, all green balls are mutually identical] Number of ways to choose zero or more black balls = (7 + 1) [since, all black balls are mutually identical] Hence, number of ways to choose zero or more balls of any colour = (10 + 1) (9 + 1) (7 + 1) Also, number of ways to choose a total of zero balls = 1 Hence, the number, if ways to choose at least one ball (irrespective of any colour) = (10 + 1) (9 + 1) (7 + 1) $-$ 1 = 879

Q22

Let

{T_n}

be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If

{T_{n + 1}} - {T_n}

= 10, then the value of n is :

A 7

B 5

C 10

D 8

Correct Answer

Option B

Solution

Number of possible triangle using n vertices = nC3 $\therefore$ Tn = nC3 then Tn + 1 = n + 1C3 Given,

{T_{n + 1}} - {T_n}

= 10 $\Rightarrow$ n + 1C3 - nC3 = 10 $\Rightarrow$

{{\left( {n + 1} \right)n\left( {n - 1} \right)} \over 6} - {{n\left( {n - 1} \right)\left( {n - 2} \right)} \over 6}

= 10 $\Rightarrow$ 3n(n - 1) = 60 $\Rightarrow$ n(n - 1) = 20 $\Rightarrow$ n2 - n - 20 = 0 $\Rightarrow$ (n - 5)(n + 4) = 0 $\therefore$ n = 5

Q23

Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A

\times

B having 3 or more elements is :

A 219

B 211

C 256

D 220

Correct Answer

Option A

Solution

A $\times$ B will have 2 $\times$ 4 = 8 elements.

The number of subsets having atleast 3 elements = 8C3 + 8C4 + 8C5 + 8C6 + 8C7 + 8C8 = 28 – (8C0 + 8C1 + 8C2) = 256 – 1 – 8 – 28 = 219

Q24

The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is:

A 120

B 72

C 216

D 192

Correct Answer

Option D

Solution

For a four digit number the first place can be filled in 3 ways with 6 or 7 or 8 and the remaining four places in 4!

ways i.e., 3 $\times$ 4!

= 72.

For a five digit number it can be arranged in 5!

ways, $\therefore$ total number of integers = (72 + 120) = 192.

Q25

If all the words (with or without meaning) having five letters,formed using the letters of the word SMALL and arranged as in a dictionary, then the position of the word SMALL is :

A

{46^{th}}

B

{59^{th}}

C

{52^{nd}}

D

{58^{th}}

Correct Answer

Option D

Solution

Clearly, number of words start with

A = {{4!} \over {2!}} = 12

Number of words start with

L = 4! = 24

Number of words start with

M = {{4!} \over {2!}} = 12

Number of words start with

SA = {{3!} \over {2!}} = 3

Number of words start with

SL = 3! = 6

Note that, next word will be "SMALL" Hence, the position of word "SMALL" is 58th.

Q26

If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is :

A 44th

B 45th

C 46th

D 47th

Correct Answer

Option C

Solution

To find the position of the word QUEEN: $\bullet$ The number of words starting with E is 4!

= 24.

$\bullet$ The number of words starting with N is

{{4!} \over 2} = 12

. $\bullet$ The number of words starting with QE is 3! = 6. $\bullet$ Number of words starting with QN is

{{3!} \over 2} = 3

.

Therefore, the position of the word QUEEN is next to the sum, 24 + 12 + 6 + 3 = 45.

That is, the word 'QUEEN' will be on 46th position.

Q27

The number of ways in which 5 boys and 3 girls can be seated on a round table if a particular boy B1 and a particular girl G1 never sit adjacent to each other, is :

A 5

\times

6!

B 6

\times

6!

C 7!

D 5

\times

7!

Correct Answer

Option A

Solution

Number of ways = Total - when B1 and G1 sit together Total ways to seat 8 people on round table = (8 - 1)!

= 7!

When B1 and G1 sit together then assume B1 and G1 are one people, so total 7 people are there and among B1 and G1 they can sit 2!

ways.

So total no of ways when B1 and G1 sit together = (7 - 1)!

$\times$ 2!

= 6!

$\times$ 2!

Number of ways = 7! - 6!

$\times$ 2!

= 6!

$\times$ (7 - 2) = 5 $\times$ 6!

Q28

A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is:

A 468

B 469

C 484

D 485

Correct Answer

Option D

Solution

X(7 Friends) Y(7 Friends) 4 Ladies 3 Men 3 Ladies 4 Men Case 1 3 0 0 3 Case 2 0 3 3 0 Case 3 2 1 1 2 Case 4 1 2 2 1 In Case 1, Case 2, Case 3 and Case 4, total 6 friends are present and 3 from X and 3 from Y and among those 6 friend 3 are ladies and 3 are men in every case.

$\therefore$ No of ways 6 friends can be invited =

({}^4{C_3} \times {}^3{C_0} \times {}^3{C_0} \times {}^4{C_3})

+

({}^4{C_0} \times {}^3{C_3} \times {}^3{C_3} \times {}^4{C_0})

+

\left( {{}^4{C_2} \times {}^3{C_1} \times {}^3{C_1} \times {}^4{C_2}} \right)

+

\left( {{}^4{C_1} \times {}^3{C_2} \times {}^3{C_2} \times {}^4{C_1}} \right)

= 16 + 1 + 324 + 144 = 485

Q29

The number of four letter words that can be formed using the letters of the word BARRACK is :

A 120

B 144

C 264

D 270

Correct Answer

Option D

Solution

Case 1 : When all the four letters different then no of words = 5C4 $\times$ 4!

Case 2 : When out of four letters two letters are R and other two different letters are chosen from B, A, C, K then the no of words = 4C2 $\times$

{{4!} \over {2!}}

= 72 Case 3 : When out of four letters two letters are A and other two different letters are chosen from B, R, C, K then the no of words = 4C2 $\times$

{{4!} \over {2!}}

= 72 Case 4 : When word is formed using two R and two A then number of words =

{{4!} \over {2!2!}}

= 6 So, total number of 4 letters words possible = 120 + 72 + 72 + 6 = 270

Q30

n

-

digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is :

A 6

B 7

C 8

D 9

Correct Answer

Option B

Solution

In n digit number first place can be filled with any one of 2, 5, 7.

So no of ways first digit can be filled = 3 Similarly, no of ways 2nd digit can be filled = 3 ways . . . . - - - - - - nth - - - - - - - = 3 ways $\therefore$ Total numbers = 3 $\times$ 3 $\times$ 3 .... n times = 3n $\therefore$ According to question, for smallest value of n, 3n > 900 36 = 729 < 900 37 = 2187 > 900 $\therefore$ n = 7