Question

C. A party of 6 is to be formed from 10 men and 7 women so as to include 3 men and 3
women. In how many ways the party can be formed if two particular women refuse to
join it?
(a) 4,200
(b) 600
(c) 3,600
(d) None

Answer 1

Step-by-step explanation:

Members required =6

Men=10

Women=7

Men needed=3

Women needed=3

2 women refused,so remaining women=7–2=5.

So, we will use combination method there

Hence Required number of ways =

10c_{3} \times 5c_{3}10c

3

×5c

3

\frac{10!}{(10 - 3)! \times 3!} \: \times \frac{5!}{(5 - 3)! \times 3!}

(10−3)!×3!

10!

×

(5−3)!×3!

5!

120 \times 10120×10

12001200