Question

in a club there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different possibilities are there?

Answer 1

The method for calculating how many arrangements are feasible when only a few items are chosen from a collection of items that are unique is as follows:

$^nC_k=\frac{n!}{k!(n-k)!}$

Where:

n – the sum of all the elements of a set

k – the number of selected objects (the order of the object may change)

! – factorial

For instance, 3!=1×2×3

So, according to the problem, there are $^7C_3$ possible groups of 3 women out of 7 women. And $^5C_2$ possible groups of 2 men out of 5 men.

Combining we get that there are $^7C_3 ~^5C2$ groups of 3 women and 2 men out of 7 women and 5 men.

Therefore, there are $^7C_3 ~^5C2=\frac{7!}{3!4!}\frac{5!}{2!3!}= 350$ possibilities.

Learn more about combinations here-

brainly.com/question/13387529

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