Question

Amit has five friends: 3 girls and 2 boys. Amit’s wife also has 5 friends: 3 boys and 2 girls. In how many maximum numbers of different ways can they invite 2 boys and 2 girls such that two of them are Amit’s friends and two are his wife’s?
(a)24
(b)38
(c)46
(d)58

Answer 1

Hint: A combination is the number of possible arrangements in a collection of items. The formula for combinations is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ , where n represents the number of items, and r represents the number of items being chosen at a time.

Complete step-by-step answer:
Given that,
Amit has five friends: 3 girls and 2 boys and Amit’s wife also has 5 friends: 3 boys and 2 girls.
Here, we want to find maximum numbers of different ways they can invite 2 boys and 2 girls such that two of them are Amit’s friends and two are his wife’s.

There can be three cases.

Case I - One boy & one girl from Amit friend and one boy and one girl from his wife friend.
Number of ways $={}^{2}{{C}_{1}}\times {}^{3}{{C}_{1}}\times {}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}}=36$

Case II- Two girls from Amit friend and two boys from his wife friend.

Number of ways \[={}^{3}{{C}_{2}}\times {}^{3}{{C}_{2}}=9\]

Case III- Two boys from Amit friend and three girls from his wife friend

Number of ways \[={}^{2}{{C}_{2}}\times {}^{2}{{C}_{2}}=1\]

Total number of ways = 36 + 9 + 1 = 46.

Hence maximum numbers of different ways are 46.

Therefore, the correct option for the given question is option (c).

Note: The difference between combinations and permutations is ordering. With permutations you care about the order of the elements, whereas with combinations you do not care about the order of the elements.