Permutations and Combinations — Page 4 | JEE Mathematics

Q31

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is :

A at least 750 but less than 1000

B at least 1000

C less than 500

D at least 500 but less than 750

Correct Answer

Option B

Solution

From 6 different novels 4 novels can be chosen =

{}^6{C_4}

ways And from 3 different dictionaries 1 can be chosen =

{}^3{C_1}

ways $\therefore$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen =

{}^6{C_4} \times {}^3{C_1}

ways Let 4 novels are N1, N2, N3, N4 and 1 dictionary is D1.

Dictionary should be in the middle.

So the arrangement will be like this _ _ D1 _ _ On those 4 blank places 4 novels N1, N2, N3, N4 can be placed.

And 4 novels can be arrange

4!

ways. $\therefore$ Total no of ways =

{}^6{C_4} \times {}^3{C_1}

\times 4!

= 1080

Q32

A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to :

A 24

B 25

C 27

D 28

Correct Answer

Option B

Solution

Given that 5 Boy, n girls. (1B, 2G) + (2B, 1G)

{}^5{C_1}.{}^n{C_2} + {}^5{C_2}.{}^n{C_1} = 1750

\Rightarrow 5.{{n\left( {n - 1} \right)} \over 2} + 10.n = 1750

\Rightarrow {{n\left( {n - 1} \right)} \over 2} + 2n = 350

\Rightarrow {n^2} - n + 4n = 700

\Rightarrow {n^2} + 3n - 700 = 0

\Rightarrow (n + 28)(n - 25) = 0

\Rightarrow n = 25, -28

Q33

The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is :

A 220 - 1

B 220

C 220 + 1

D 221

Correct Answer

Option B

Solution

21C0 + 21C1 + 21C2 + ....... + 21C10 =

{{{2^{21}}} \over 2} = {2^{20}}

Q34

The number of 6 digit numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no digit is repeated is :

A 36

B 60

C 72

D 48

Correct Answer

Option B

Solution

digit 0, 1, 2, 5, 7, 9

\left( {{a_1} + {a_3} + {a_5}} \right) - \left( {{a_2} + {a_4} + {a_6}} \right) = 11\,K

so (1, 2, 9) (0, 5, 7) Now number of ways to arranging them = 3! × 3! + 3! × 2 × 2 = 6 × 6 + 6 × 4 = 6 × 10 = 60

Q35

A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then :

A n = m – 8

B m = n = 78

C m + n = 68

D m = n = 68

Correct Answer

Option B

Solution

At least 6 males means in the committee there can be 6 males or 7 males or 8 males. $\therefore$ m =

{}^8{C_6} \times {}^5{C_5} + {}^8{C_7} \times {}^5{C_4} + {}^8{C_8} \times {}^5{C_3}

= 78 At least 3 females means in the committee there can be 3 females or 4 females or 5 females. $\therefore$ n =

{}^5{C_3} \times {}^8{C_8} + {}^5{C_4} \times {}^8{C_7} + {}^5{C_5} \times {}^8{C_6}

= 78 So, m = n = 78

Q36

All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is :

A 175

B 162

C 160

D 180

Correct Answer

Option D

Solution

For those three odd digit numbers 1, 1, 3 we can choose any three positions out of the four even positions.

$\therefore$ No of ways we can choose 3 positions out of the 4 positions = 4C3 After choosing those three positions, number of ways we can arrange three odd digit numbers = 4C3 $\times$

{{3!} \over {2!}}

Then the remaining 6 digits can be arrange =

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

ways $\therefore$ Total number of 9 digit numbers = 4C3 $\times$

{{3!} \over {2!}}

$\times$

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

= 180

Q37

There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is :

A 12

B 9

C 7

D 11

Correct Answer

Option A

Solution

Let m-men, 2-women mC2 $\times$ 2 = mC1 2C1 . 2 + 84 m2 $-$ 5m $-$ 84 = 0 $\Rightarrow$ (m $-$ 12) (m + 7) = 0 m = 12

Q38

Consider three boxes, each containing, 10 balls labelled 1, 2, … , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni, the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is :

A 164

B 240

C 82

D 120

Correct Answer

Option D

Solution

Number of ways = 10C3 = 120

Q39

The number of natural numbers less than 7,000 which can be formed by using the digits 0, 1, 3, 7, 9 (repitition of digits allowed) is equal to :

A 374

B 372

C 375

D 250

Correct Answer

Option A

Solution

Total no 1 digit numbers possible = 4 (allowed digits 1, 3, 7, 9) Total no 2 digit numbers possible = 4 $\times$ 5 = 20 Total no 3 digit numbers possible = 4 $\times$ 5 $\times$ 5 = 100 Total no 4 digit numbers possible = 2 $\times$ 5 $\times$ 5 $\times$ 5 = 250 So the number of natural numbers less than 7,000 possible are = 4 + 20 + 100 + 250 = 374

Q40

Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?

A 2! 3! 4!

B (3!)3.(4!)

C 3! (4!)3

D (3!)2.(4!)

Correct Answer

Option B

Solution

F1 $\to$ 3 members F2 $\to$ 3 members F3 $\to$ 4 members Total arrangements of three families = 3!

Arrangement between members of F1 family = 3!

Arrangement between members of F2 family = 3!

Arrangement between members of F3 family = 4!

$\therefore$ Total numbers of ways can they be seated so that the same family members are not separated = 3!

$\times$ 3!

$\times$ 4!

= (3!)

3.(

4!)