(1 point) suppose that a department contains 8 men and 17 women. how many different committees of 6 members are possible if the committee must have strictly more women than men?

There are three cases that we can solve.

CASE 1:

Committee with 6 women (no men)

Choose the 6 women --> C (17,6)

= (17 x 16 x 15 x 14 x 13 x 12) / (6 x 5 x 4 x 3 x 2 x 1)

= 12,376 ways

CASE 2:

Committee with 5 women and 1 man

Choose the 5 women --> C(17,5)

= 6,188 ways

Choose the 1 man --> C(8,1)

= 8 ways

Multiply:

C(17,5) * C(8,1)

=6,188 x 8

= 49,504 ways

CASE 3:

Committee with 4 women and 2 men:

Choose the 4 women --> C(17,4)

= 2,380 ways

Choose the 2 men --> C(8,2)

= 28 ways

Multiply:

C(17,4) * C(8,2)

=2,380 x 28

= 66,640 ways

There aren't any other valid cases (e.g. 3W+3M, 2W+4M, 1W+5M, 6M) so you are done.

Add up all the valid cases:

12,376 + 49,504 + 66,640 = 128,520 ways

There are 128,520 ways to form a committee of 6 members (where there are strictly more women than men)

think about it deeply and passionately about this property

i am sorry

the answer would be c:   ax + b = ax + c

step-by-step explanation: