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The Question

Does the following inequality hold: $$n!k! \le (n+k-1)! \quad \text{for positive integers} \quad n \ge 1,\\ k \ge 1 \text{.}$$ I suspect it does but I'm at a loss on how to prove (or disprove) this.

What I've tried

I've been able to show that the simpler inequality: $ n!k! \le (n+k)!\\ $ holds by dividing both sides by $k!$ so that $n$-terms remain on both sides: $ \prod_{p=1}^{n} p \le \prod_{p=1}^{n} (p+k) $ which is clearly true for $k \ge 1$.

Applying the same process for the original inequality yields $ \prod_{p=1}^{n} p \le \prod_{p=1}^{n-1} (p+k) $ which feels inconclusive due to the right-hand-side having one fewer terms than the left-hand-side.

I don't have much of a background in combinatorics so any help would be appreciated.

Martin Brandenburg
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xcelsior
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    Hint: For fixed $n$, use induction on $k$. – Feng Jan 04 '24 at 01:17
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    A warning, a first inclination might have been to try to claim that $n!k!$ is always a divisor of $(n+k-1)!$ (and thus smaller) by way of some combinatorial argument that $\binom{n+k}{n}\cdot \frac{1}{n+k}$ should have been some answer to a counting problem, however that doesn't work, noting as a counterexample something like when $n=k=p$ for some prime $p$, the factor of $p$ will have appeared in the denominator thrice when expanded while only appearing in the numerator twice. – JMoravitz Jan 04 '24 at 01:26

4 Answers4

9

Consider the distinct subsets $A_1 = \{1, \ldots, n\}$, $A_{2} = \{2, \ldots, n+1\}, \ldots,$ $A_{k} = \{k, \ldots, n+k-1\}$ of size $n$.

It follows that $k \leq \binom{n+k-1}{n}$. Rearranging yields the original inequality.

2

By writing $N=n+k$ the claim is equivalent to $\binom{N}{n} \geq N$. To prove this we may assume $n \leq N/2$, and then use that these binomial coefficients $\binom{N}{1},\binom{N}{2},\dotsc$ are in fact non-decreasing, starting with $\binom{N}{1}=N$ (see here, you can directly verify $\binom{N}{n-1} < \binom{N}{n}$).

Martin Brandenburg
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2

For arbitrarily fixed $n\geq 1$, we argue by induction on $k$. For $k=1$, the desired inequality holds trivially. Assume that the inequality holds for some $k\geq1$, i.e., $$n!k! \le (n+k-1)!\tag{$*$},$$ we prove that it holds also for $k+1$, hence our aim is to show that $$n!(k+1)!\le(n+k)!\tag{$**$}.$$ By our induction assumption $(*)$, we have $$n!(k+1)!=n!k!(k+1)\le (n+k-1)!(k+1).$$ To show $(**)$, it suffices to prove that $(n+k-1)!(k+1)\le(n+k)!$, which is equivalent to $k+1\leq n+k$, and this follows from $n\geq 1$. Hence $(**)$ holds and thus the inequality is porved by induction.

Feng
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1

A combinatorial proof here: $n-1$ adults and $k$ children stand in a line.

Scenario 1

They can stand in any order they want. The number of possibilities is $(n+k-1)!$

Scenario 2

There are no adults between any two children. The children can be arranged in $k!$ ways. Subsequently, the $n-1$ adults and $1$ group of children can be arranged in $n!$ ways. Therefore the number of possibilities is $n!k!$.

Conclusion

Clearly the number of possibilities in scenario 1 is greater or equal to that in scenario 2. Therefore we obtain what we need:

$$ (n+k-1)!\geq n! k! $$

acat3
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