Groomsmen and Bridesmaids

A couple wants to invite their friends to be at their wedding party. The party will consist of five groomsmen and five bridesmaids. The groom has eight possible groomsmen, and the bride has 11 bridesmaids.

1. How many groups are possible?

The order doesn’t matter, so it’s combinations.

$_nC_r = \frac{n!}{(n-r)!r!}$

For the groom, it becomes,

$_8C_5 = \frac{8!}{(8-5)!5!} = \frac{8*7*6*5*4*3*2*1}{(3*2*1)*(5*4*3*2*1)} = 8*7 = 56$

For the bride,

$_{11}C_5 = \frac{11!}{(11-5)!5!} = \frac{11*10*9*8*7*6*5*4*3*2*1}{(6*5*4*3*2*1)*(5*4*3*2*1)} = 11*3*2*7 = 462$

And the overall combinations are: 56 x 462 = 25872

choose(8,5)*choose(11,5)

2. Suppose one possible groomsman and one possible bridesmaid refuse to be together; how many groups are possible?

First, we leave those and make groups:

choose(7,5)*choose(10,5)

Now, add the situation where that member from one side is present, and the one from the other is moving out (there are two instances).

choose(1,1)*choose(7,4)*choose(10,5) 
choose(1,1)*choose(7,5)*choose(10,4)

8820
4410

Sum all up:

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