# Number of permutations when composing recital

• Mar 6th 2012, 07:45 AM
timmeh
Number of permutations when composing recital
The problem is:

A pianist wants to prepare a recital that will consist of 3 Classical, 4 Romantic and 2 Contemporary pieces.
In how many ways can he order the piano pieces if he wants to play all the pieces of a genre together?

I'm a bit confused with these kind of problems, when the first choice could influence a second choice. If the first choice would be a Classical piece, then the following pieces must be Classical as well. Anyway, my reasoning was as follows, using the fundamental counting principle:

In total there are 9 "events (E's)", because there are 9 pieces of music in total.

E1 = 9 choices
E2 = 2 choices, if for example, the first choice (E1) was a Classical piece
E3 = then must be another Classical piece which is 1 remaining choice
E4 = 6 choices (4R + 2C)
E5 = if Romantic was chosen at E4, then here are 3 choices
E6 = 2 choices
E7 = 1 choice
E8 = 2 choices, only Contemporary left
E9 = 1 choice

So the total number of ways this can happen is: 9*2*1*6*3*2*1*2*1 = 1260 ways.

According to the answer sheet this is incorrect. The correct answer should be 1728 ways.

Can somebody show me where my reasoning if faulty?
• Mar 6th 2012, 07:51 AM
timmeh
Re: Number of permutations when composing recital
Magically after writing my problem I think I found the right reasoning.

You can order the Classical pieces in 3*2*1 = 6 ways
The Romantic pieces in 4*3*2*1 = 24 ways
The Contemporary pieces in 2*1 = 2 ways

You can order the genres in 3*2*1 = 6 ways

Total number of ways = 6*24*2*6 = 1728 ways

Right!? :P
• Mar 6th 2012, 08:37 AM
Soroban
Re: Number of permutations when composing recital

Right!