Here's my problem: The post office has 15 types of stamps and only 2 of them are royals. If I purchase 3 stamps, how many ways can I come home with at least 1 royal stamp?
The answer I was given is 325 ways. I can't figure out how to get that.
I thought, find out how many ways to get any 3 stamps from 15, then subtract from that the number of ways to get 3 non-royal stamps from 13, but that gives an answer of 169. Somebody please help!
On
There are 15 types of stamps. The number of multisets consisting of 3 stamps is $\binom{15+3-1}{3}$. (Note that we are allowed to buy more than one of the same type of stamp.) The number of multisets consisting of 3 stamps, none of which are royals, is $\binom{13+3-1}{3}$. Thus, the number of multisets consisting of 3 stamps, such that at least one of the stamps is a royal, is $$ \binom{15+3-1}{3} - \binom{13+3-1}{3} = 225. $$
I bet there's a typo in the answer you were given, and it was supposed to say $225$.
First suppose the $3$ stamps purchased are distinct types . . .
The number of subsets of $3$ stamps with exactly one royal is $$\binom{2}{1}\binom{13}{2}=156$$ The number of subsets of $3$ stamps with two royals is $$\binom{2}{2}\binom{13}{1}=13$$ Summing the results gives $156+13=169$.
But the problem doesn't specify that the $3$ stamps are distinct types, so that allows for more possibilities . . .
If all three stamps are the same type, that yields two more qualifying results.
So now the count is $169+2=171$.
If exactly two are the same type, we have some cases . . .
If the two same-type ones are royals, and the third one is also royal, the count is $$\binom{2}{1}\binom{1}{1}=2$$ If the two same-type ones are royals, and the third one is non-royal, the count is $$\binom{2}{1}\binom{13}{1}=26$$ If the two same-type ones are non-royals, the count is $$\binom{13}{1}\binom{2}{1}=26$$ Summing the results gives $171+2+26+26=225$.