MMM #24 Winner

The winner for this edition of MMM is Jonathan. I was just playing Rock Band 2 over the weekend and I must say it’s one of my top 5 favorite games of all time. If anyone wants to play Rock Band with me online and owns a PS3, let me know! We will surely rock.

Anyhow, the number of submissions were down for this contest, so I can only conclude that the problem was either too hard, or not interesting. It could have been a combination of both. You can bet that the next problem we host at blinkdagger will be a good one!

Next week’s MMM, hosted at wildaboutmath.com should also be a good one, so don’t miss out! I got a sneak peak from Sol, and I can say that it was not an easy problem! Good luck to all.

Now, onto the answer!

The Answer by Vladimir

The total number of 8-title-programs is N=5^8=390625. Let’s count the number M of “good” programs.
If blinkdagger have played all 5 songs then there are 3 different situations:

a). They played 4 songs and 4 times the last one (for example [1 2 3 4 5 1 1 1])
There are C_5^1=5 different song lists (i.e. [1 1 1 1 2 3 4 5], [1 2 2 2 2 3 4 5], [1 2 3 3 3 3 4 5], [1 2 3 4 4 4 4 5], [1 2 3 4 5 5 5 5])
Number of displacements of songs in each program is 8!/4! (total number of permutations divided by number of permutations of equal elements).
Therefore there are M_1=5*8!/4!=8400 such situations.

b). They played once 3 songs, 3 times one another and 2 times the last one (for example [1 2 3 4 5 1 1 2])
There are 2*C_5^2=2*10=20 such song lists (C_5^2 - number of selections of 2 songs from 5 and we have to multiply by 2, because there are 2 variants of their displacement, for instance [1 1 2] and [1 2 2]).
Number of displacements of songs in each such program is 8!/(3!*2!).
Therefore there are M_2=20*8!/(3!*2!)=67200 such situations.

c). They played all 5 songs and another 3 different songs.
There are C_5^3=10 such lists.
Number of displacements of songs in each such program is 8!/(2!*2!*2!).
Therefore there are M_3=10*8!/(2!*2!*2!)=50400 such situations.

Summarizing there are M=M_1+M_2+M_3=126000 “good” programs, and the probability is M/N=0.32256.