There are 13 basketball players. All the players sit in a row on a bench. Determine the number of ways to arrange them such that the two youngest players sit together.
Suppose the 13 basketball players are
A, B, C, D, E, F, G, H, I, J, K, L, M
Suppose L and M are the youngest. Then we have to arrange either these
12 "things", where a "thing" is either a player or the pair of youngest
players, where L is left of M
A, B, C, D, E, F, G, H, I, J, K, (LM)
or these 12 "things" where L is right of M
A, B, C, D, E, F, G, H, I, J, K, (ML)
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In each case you can choose the left most thing 12 ways, the next one
11 ways, the next one 10, ways,... the next to the right one 2 ways and the
thing on the far right 1 way. That's
12×11×10×9×8×7×6×5×4×3×2×1 = 12! = 479001600
But we must double this since we can have (LM) or (ML)
Answer = 479001600×2 = 958003200
Looks like you're exactly right!
Edwin
