Combinations.

Is the following necessarily true?

n ! ( r 1 ! ) ( r 2 ! ) ( r 3 ! ) . . . N \large \frac{n!}{(r_{1}!)(r_{2}!)(r_{3}!)...} \in \mathbb{N}

where r 1 + r 2 + r 3 . . . n r_{1} + r_{2} + r_{3}... \leq n

Note: 0 is not an element of the natural numbers set.

Bonus: Proofs are encouraged.

Yes No

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1 solution

Raghu Raman Ravi
Sep 30, 2017

Let N = i = 1 k ( n j = 1 i 1 r j r i ) \large N=\prod_{i=1}^{k}\binom{n-\sum_{j=1}^{i-1}r_{j}}{r_{i}} Where i = 1 k r i n \large\sum_{i=1}^{k}r_{i}\leq n

Since N is a product of Binomial coefficients, it is a positive integer. But, N = i = 1 k ( n j = 1 i 1 r j ) ! ( r i ! ) ( n j = 1 i r j ) ! \large N=\prod_{i=1}^{k}\frac{(n-\sum_{j=1}^{i-1}r_{j})!}{(r_{i}!)(n-\sum_{j=1}^{i}r_{j})!} = n ! r 1 ! ( n r 1 ) ! × ( n r 1 ) ! r 2 ! ( n r 1 r 2 ) ! . . . . . . . . \large =\frac{n!}{r_{1}!(n-r_{1})!}\times\frac{(n-r_{1})!}{r_{2}!(n-r_{1}-r_{2})!}........ = n ! r 1 ! r 2 ! . . . . r k ! ( n j = 1 k r j ) ! \large =\frac{n!}{r_{1}!r_{2}!....r_{k}!(n-\sum_{j=1}^{k}r_{j})!} Therefore, The expression given in the question is N × ( n j = 1 k r j ) ! \large N\times(n-\sum_{j=1}^{k}r_{j})! Which is an integer.

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