There are 3 positions. There are 6 boys and 5 girls. Total of 11
a) no restrictions
This is a permutation ;
nPr ="\\frac {n!} {(n-r)!}"
="\\frac {11!}{(11-3)!}"
=990 ways
b) at least one boy chosen.
This is where 1, 2 or 3 boys can be selected.
We subtract when no boy(all girls) is selected from when no restriction is put in place.
11P3-5P3
="\\frac {11!}{(11-3)!}-\\frac {5!}{(5-3)!}\u200b"
=990- 60
=930 ways
c) only one girl chosen
This means that 2 boys are chosen
And there are 3 positions.
5P1*6P2*3
="\\frac {5!}{(5-1)!}*\\frac {6!}{(6-2)!}*3"
=450 ways
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