Permutations and Combinations Questions — JEE Mathematics

Q1

Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are :

A 216

B 375

C 400

D 720

Correct Answer

Option D

Solution

$\therefore$ Total no of ways = 5 $\times$ 6 $\times$ 6 $\times$

{}^4{C_1}

= 720

Q2

The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1, 3, 5, 7, 9 without repetition, is :

A 120

B 6

C 72

D 12

Correct Answer

Option C

Solution

Numbers between 5000 & 10000 Using digits 1, 3, 5, 7, 9 Total Numbers = 3 × 4 × 3 × 2 = 72

Q3

Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is :

A

{5 \over 2}\left( {6!} \right)

B

{6!}

C 56

D

{1 \over 2}\left( {6!} \right)

Correct Answer

Option A

Solution

Here none number repeats more than once.

We can choose the number which repeats more than once among 1, 3, 5, 7, 9 in 5C1 ways.

Let number 3 repeats more than once.

So six digits are 1, 3, 3, 5, 7, 9.

We can arrange those six digits in

{{6!} \over {2!}}

ways. $\therefore$ Total six digit numbers = 5C1 $\times$

{{6!} \over {2!}}

=

{5 \over 2}\left( {6!} \right)

Q4

The number of ordered pairs (r, k) for which 6.35Cr = (k2 - 3). 36Cr + 1, where k is an integer, is :

A 6

B 3

C 2

D 4

Correct Answer

Option D

Solution

6.35Cr = (k2 - 3). 36Cr + 1 $\Rightarrow$ 6.35Cr = (k2 - 3).

{{36} \over {r + 1}}

35Cr $\Rightarrow$ k2 - 3 =

{{r + 1} \over 6}

Possible values of r for integral values of k, are r = 5, 35 number of ordered pairs are 4 (5, 2), (5, –2), (35, 3), (35, 3)

Q5

The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by

A

7!\, \times 5!\,\,

B

6!\, \times 5!

C

30!

D

5!\, \times 4!

Correct Answer

Option B

Solution

6 men can sit at the round table =

\left( {6 - 1} \right)! = 5!

ways Now at the round table among 6 men there are 6 empty places and 5 women can sit at those 6 empty positions.

So total no of ways 6 men and 5 women can dine at the round table =

5!\, \times {}^6{C_5} \times 5!

=

5!\, \times 6 \times 5!

=

5!\, \times 6!

Q6

The number of four-digit numbers strictly greater than 4321 that can be formed using the digits 0,1,2,3,4,5 (repetition of digits is allowed) is :

A 306

B 288

C 310

D 360

Correct Answer

Option C

Solution

So total 4 digit numbers possibe = 216 + 72 + 18 + 4 = 310

Q7

How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order

A 480

B 240

C 360

D 120

Correct Answer

Option C

Solution

In the word ''GARDEN'', there are two vowels A and E present, and A should come always before E.

\therefore\,\,\,

Total no of ways =

{{6!} \over {2!}}

= 360 Here A and E has fixed order that is why we divide by 2!.

Q8

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

A

{}^8{C_3}

B 21

C

{3^8}

D 5

Correct Answer

Option B

Solution

To distribute n objects among p people where everyone should get atleast one object, then number of ways to distribute those n objects =

{}^{n - 1}{C_{p - 1}}

For this question, n = 8 and p = 3 $\therefore$ Number of ways =

{}^{8 - 1}{C_{3 - 1}}

=

{}^7{C_2}

= 21

Q9

The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only is :

A 35

B 42

C 82

D 77

Correct Answer

Option D

Solution

(I) First possibility is 1, 1, 1, 1, 1, 2, 3 required number =

{{7!} \over {5!}}

= 7 $\times$ 6 = 42 (II) Second possibility is 1, 1, 1, 1, 2, 2, 2 required number =

{{7!} \over {4!3!}} = {{7 \times 6 \times 5} \over 6} = 35

Total = 42 + 35 = 77

Q10

Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total number of beams is :

A 180

B 210

C 170

D 190

Correct Answer

Option C

Solution

Any two non-adjacent pillers are joined by beams $\therefore$ number of beams = number of diagonals = 20C2 - 20 = 170