1. ## Combinations

My answer to this question is 2 X 5! = 240. The book gives 1440, presumably 2 X 6!.

A concert pianist has prepared seven different pieces of music for recital, three of which are modern and four are classical. Calculate the number of different orders in which she can play the seven pieces when the recital must start and end with classical pieces.

2. Originally Posted by Stuck Man
My answer to this question is 2 X 5! = 240. The book gives 1440, presumably 2 X 6!.

A concert pianist has prepared seven different pieces of music for recital, three of which are modern and four are classical. Calculate the number of different orders in which she can play the seven pieces when the recital must start and end with classical pieces.
I get there via 4*(6!-3*5!), alternatively, 4*3*5!, alternatively, P(4,2)*5!

3. P(4,2)*5! is a valid method. The end pieces are the permutations of any 2 from 4 classical pieces. The centre pieces are the permutations of the 5 remaining pieces.

4. Originally Posted by Stuck Man
P(4,2)*5! is a valid method. The end pieces are the permutations of any 2 from 4 classical pieces. The centre pieces are the permutations of the 5 remaining pieces.
Yes and the other two methods are also valid.

4*(6!-3*5!) -- 4 ways to choose the first piece, then 6! permutations of the rest, but 3*5! of those don't end with classical.

4*3*5! -- 4 ways to choose first piece, 3 ways to choose last piece, 5! ways to permute the rest.

5. Alternatively, the central 5 contain $\binom{4}{2}$ pairs of classical pieces, along with the 3 modern pieces.

There are 5! arrangements of the central 5 and for each of these, there are 2! arrangements
of the first and last classical pieces.

That's $\binom{4}{2}5!2!$

Or...

There are $\binom{4}{2}$ ways to select the 1st and last classical pieces.

There are 5! arrangements of the central 5 pieces and for each of these there are 2! arrangements
of the outer 2 pieces....again $\binom{4}{2}5!2!$