combinatorial argument
2009-01-22 5:43 am
Provide a combinatorial argument to show that if n and k are positive integers with n = 3k, then n!/(3!)^k is an integer
回答 (1)
2009-01-22 6:31 am
✔ 最佳答案
Consider the following problem: A set of n=3k distinct items os to be divided into k distinct groupsof sizes 3,3,3,3,3...3 (k times) where 3k=n How many different divisions are possible ? It can be shown that the value is n!/(3!)^k (using nCr and multiplication rule) and so n!/(3!)^k is an integer
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