(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)

part a

in a city ice cream is mostly only bought when the temperature is very high as ice cream cools down our

so we can derive the conclusion that as the temperature increases the number of ice creams sold also increases

part b

y intercept is 5

slope is 2

happy to

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