Different answers when I split a combination question into cases or solve in one step. The students producing a school fashion show plan to have four pieces of music played. The music students have come up with 18 pieces: 6 for piano, 5 for recorder, and 7 for guitar. The students want to use at least 1 piece for the piano. In how many ways can the group choose the 4 pieces of music? Explain your reasoning.
I am not actually sure of the answer to this question. My solution was $$6C1(12C3)+6C2(12C2)+6C3(12)+6C4=2565$$
Then I thought of another solution which I thought should also work.
$$6C1(17C3)=4080$$
The first solution is split into cases of choosing 1,2,3,4 piano pieces. The second solution chooses 1 piano piece first and then selects three pieces of music from the rest. Which solution is right? And why do the two solutions not yield the same answer? Thank you.
 A: The first answer is correct.
Your second answer overcounts those selections in which you choose more than one piano piece.  For instance, if you choose the first and second piano pieces, you have counted them twice, once when you select the first piano piece and then select the second piece as one of the other three pieces and once when you select the second piano piece and select the first piece as one of the other three pieces.  If three piano pieces are selected you count each one three times; if four piano pieces are selected you count each one four times.  Notice that 
$$\binom{6}{1}\binom{12}{3} + 2\binom{6}{2}\binom{12}{2} + 3\binom{6}{3}\binom{12}{1} + 4\binom{6}{4} = 4080$$
A: Another approach you could use is to count the number of ways to choose any 4 pieces,  
and then subtract the ways to choose pieces which do not involve the piano.
This gives $\dbinom{18}{4}-\dbinom{12}{4}=3060-495=2565$
A: I would go with casework, as we would have no overlapping counts. Those can get quite annoying. The following summation does this, for $1$ to $4$ piano pieces. $$\sum_{n = 1}^{4}\dbinom{6}{n} \cdot \dbinom{12}{4 - n} = \boxed{2565}.$$
