In how many ways can you permute four 1's and fifty 2's such that at least five 2's lie between each pair of 1's?
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After placing 5 2 's in each of the 3 gaps between the 4 1 's, we have 3 5 2 's left to place, in any combination, in the 5 available positions, namely before the first 1 , in the 3 gaps between the 1 's, and after the last 1 . This is a 'stars and bars' calculation with solution ( 3 5 3 5 + 5 − 1 ) = 8 2 2 5 1 combinations.