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ABCDEABCDE is one (Posted on 2006-05-17) Difficulty: 2 of 5
How many arrangements of the letters AABBCCDDEE are there such that there are no double letters in the sequence?

See The Solution Submitted by Jer    
Rating: 4.5000 (2 votes)

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Found this in the OEIS Comment 3 of 3 |
Searching the OEIS with the answer 39480 quickly finds A114938 "Number of permutations of the multiset {1,1,2,2,....,n,n} with no two consecutive terms equal."  This problem asks to compute a(5).

The OEIS provides several formulas including a recursive formula:
a(n) = n*(2*n-1)*a(n-1) + (n-1)*n*a(n-2)

a(1) = 0, trivially.  a(2) = 2, for ABAB and BABA.  Then a(5) is pretty easy to calculate by hand by stepping through the sequence: 0, 2, 30, 864, 39480, ....

  Posted by Brian Smith on 2017-01-15 11:37:44
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