Question

There are 42 men and 16 women in a party. Each man shakes his hand only with men(all) and each women shakes with all the women only.Find max no. og hand shakes?

Answer 1

Hello User this problem is based on permutation and combination.
Here Combination is used.
Maximum Number of hand shakes
= 42C2+16C2
=
$= \frac{42 \times 41}{2 \times 1} + \frac{16 \times 15}{2 \times 1 } \\ = 21 \times 41 + 8 \times 15 \\ 861 + 120 = 981$

Answer 2

Answer:

Step-by-step explanation:

Every man shakes hand with all other men = 42 C2 =42*41 / 1*2 = 861

Every woman shake hand with all other women = 16 C2 = 16 *15 / 1*2 =120

Total = 42C2+16C2 = 861+120 = 981