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Show all work. A disc jockey has 11 songs to play. Seven are slow songs, and four are fast songs.?

Show all work. A disc jockey has 11 songs to play. Seven are slow songs, and four are fast songs. Each song is to be played only once. In how many ways can the disc jockey play the 11 songs if The songs can be played in any order. The first song must be a slow song and the last song must be a slow song. The first two songs must be fast songs.

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1. 3
2. 1. The songs can be played in any order. The answer is 11! or 11x10x9x8x7x6x5x4x3x2x1 = 39916800 permutations. This is because The first song can be any of 11 The 2nd song can be any of 10 (he has already played one) The 3rd song can be any of 9 (he has already played 2) The 4th song can be any of 8 (he has already played 3) and so on 11x10x9x8x7x6x5x4x3x2x1 = 11! 2. The first and last songs must be slow The first can be any of 7 Now select the last song - it can therefore be any of the SIX remaining slow songs (put it on one side to play last - but remember it is one of SIX) That leaves nine other records that can be played in any order and they can be played in 9! different ways. Already determined that the 1st of the 11 records is any of 7 The second can be any of the residue 9 (not the first or last slow songs) The 3rd can be any of 8 The 4th can be any of 7 And so on The 9th can be any of 2 The tenth, just one (the last of the residue 9) The 11th already determined one of SIX (see above) Therefore the 11 songs can be played in 7 x 9! x 6 different permutations = 7x9x8x7x6x5x4x3x2x1x6 = 15240960 3. The first 2 songs must be fast songs The first can be any of 4 fast songs The second can be any of the 3 fast songs remaining Then the last nine can be played in any order = 9! Therefore there are 4 x 3 x 9! permutations Therefore 4x3x9x8x7x6x5x4x3x2x1 = 4354560 permutations