In how many ways can a team of 3 boys and 2 girls be selected from 6 boys and 5 girls?
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Hello, I have been reading about this stuff, too (for fun)! Let’s reverse the order of the terms in your clauses to end up with an operation (a division) that follows logically from the problem and let’s understand that we do not really need to count reciprocal results here (we only want to know the unique relationships).
BOYS:
6C3
6!/3!*(6!-3!) = 6!/3!*3!
(6*5*4*3*2*1)/(3*2*1*3*2*1) = 20 ways
GIRLS:
5C2
5!/2!*(5-2)! = 5!/2!*3!
(5*4*3*2*1) / (2*1*3*2*1) = 10 ways
The combined effort would therefore be 10*20 ways = 200 ways.
BOYS:
6C3
6!/3!*(6!-3!) = 6!/3!*3!
(6*5*4*3*2*1)/(3*2*1*3*2*1) = 20 ways
GIRLS:
5C2
5!/2!*(5-2)! = 5!/2!*3!
(5*4*3*2*1) / (2*1*3*2*1) = 10 ways
The combined effort would therefore be 10*20 ways = 200 ways.
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