Question

In how many ways can a committee of 3 Democrats and 2 Republicans be formed from a group of 11 Democrats and 10 ​Republicans?

Answer 1

Explanation:

that sounds like the sequence does not matter (e.g. the selection of Democrat A and Democrat B is the same as the selection of Democrat B and Democrat A).

and repetition is not possible (there cannot be a single person appearing twice in a selection).

so, we have combinations without repetition.

how many options do we have to select the 3 Democrats out of 11 ?

C(11, 3) = 11! / (3! × (11 - 3)!) = 11! / (3! × 8!) = 11×10×9/(3×2) =

= 11×5×3 = 165

how many options do we have to select the 2 Republicans out of 10 ?

C(10, 2) = 10! / (2! × (10 - 2)!) = 10! / (2! × 8!) = 10×9/2 =

= 5×9 = 45

how many optics now to build the committee ?

for each of the Democrat selections we have all Republican options.

so, we have

165 × 45 = 7,425

different ways to build the committee.