In how many ways can a committee
of 5 men and 3 women be selected
from 9 men and 7 women ?

Answer：

use the same method

Step-by-step explanation:

Men=5

Women=3

Available Strength:

Men=7

Women=6

Objective:Finding no of such committees where particular 2 women are never together.

So initially ,Total possible number of commitee=7c5×6c3=420.

Now we need to find total committes where two particular women say(Sita and Gita ) are never together.

Approach—->

As the commitee requires 5 men and 3 women.

Let us first find number of committees where Sita and Gita are always together.

Hence we reserve 2 women seats for them.

Now we need one women out of remaining 4 women.

So possible number of such commitees 7c5×4c1.

=84.

Hence number of committees where Sita and Gita are never together

=420–84

=336.

he kicked it up? is this the answer ?

step-by-step explanation:

need more detail

step-by-step explanation:

lele