Given a group of 8 women and 11 men, how many different ways are there of choosing one man and one woman for a committee?

Any of the 11 man can be chosen, and combined with any of the 8 women.

Assume we select man1. The selected committee can be:

(m1,w1), (m1,w2), (m1,w3), (m1,w4), (m1,w5), (m1,w6), (m1,w7), (m1,w8), 

so there are 8 committees selections with man1 in them.

we could repeat the same procedure for the remaining 10 men, and get 8 committees where each of them is a member.

so there are 11*8=88 ways of choosing 1 man and 1 woman.

88

hey cutie use pythagoras theorem u will get it

Step-by-step explanation:

x²= 9²+7²

X²= 130

now take root of 130