Q.1 Amit has five friends 3 girls & 2 boys . Amit’s wife also has 5 friends: 3 boys & 2 girls. In how many maximum no. of different ways can they invite 2 boys & 2 girls such that two of them are Amit’s friends & two are his wife’s ?

A) 38

B) 24

C) 46

D) 58

Ans. C

     AMIT’S FRIENDS      AMIT’S WIFE FRIENDS

   BOYS

(2) GIRLS

(3) BOYS

(3) GIRLS

(2) WAYS OF SELECTION

CASE I    2  0  0  2  2C2*3C0*3C0*2C2

=1*1*1*1=1

CASE II   0  2  2  0  2C0*3C2*3C2*2C0

=1*3*3*1=9

CASE III  1  1  1  1  2C1*3C1*3C1*2C1

=2*3*3*2=36

Total selection = 1+9+36 = 46

Q.2 Groups each containing 3 boys are to be formed out of 5 boys A,B,C,D & E such that no group can contain both C & D together. What is the maximum number of different groups ?

A) 5

B) 6

C) 7

D) 8

Ans. C

Maximum no. of different groups = ABC, ABD, ABE, BCE,BDE,CEA,DEA

=7

Q.3 Each of the 3 persons is to be given some identical items such that no product of the number of items received by the each of the three persons is equal to 30.In how many maximum different ways can this distribution be done ?

A) 27

B) 21

C) 24

D) 33

Ans. A

Items can be distributed in such a manner that product of the number of items is equal to 30

(3)3 = 27 ways

Q.4 Each of the two women & three men is to occupy one chair out of eight chairs , each of which numbered from 1 to 8. First women are to occupy any two chairs from those numbered 1 to 4; & then the three men would occupy any three chairs  out of the remaining six chairs. What is the maximum number of different ways in which this can be done ?

A) 1440

B) 132

C) 40

D) 3660

Ans. A

One Women can  sit in four different ways

Second women can  sit on anyone chair in three different ways

So, the different ways of sitting becomes 4*3 = 12

The ways of sitting of three men out of the remaining six chairs is 6P3

i.e 6*5*4 = 120

Therefore, total number of ways = 12*120 = 1440

Q.5 A mixed double tennis game is to be played between the two teams (Consisting of one male & one female) There are four married couples. No team is to consist of husband & his wife. What is the maximum number of games that can be played ?

A) 48

B) 36

C) 12

D) 42

Ans. A

Married couples= MF  MF  MF  MF

                             AB   CD  EF  GH

Possible Teams= AD CB EB  GB

                            AF  CF  ED  GD

Team AD can play Only with CB,CF,CH,EB,EH,GB,GF

Team AD cannot play with : AF,AH, ED & GD

This same can apply with all the teams ,

Therefore, total number of matches =12*7 = 84

But every match includes two teams , thus actual number of matches = 84/2 = 48