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Suppose you have an expression of the form $$\frac{n!}{k_1! k_2! \cdots k_r!}$$ where $\displaystyle \sum k_i \le n$. The expression must be an integer, because it's the number of permutations of $n$ objects where the $k_i$ are the number of repeated objects. For example, there are $\frac{11!}{4!4!2!}$ ways to order the letters in the word MISSISSIPPI.

The question is, can every integer in the denominator be canceled with some single integer in the numerator (i.e. every integer in the denominator should divide some integer in the numerator), or is it possible that in order to cancel some integer in the denominator you might have to combine (multiply) some number of integers in the numerator? For example, is it possible for an expression in the above form to look something like $\frac{2 \cdot 2 \cdot 3 \cdot 3}{4 \cdot 9}$ in which case you would not be able to cancel the 4 or the 9 in the denominator with any single integer in the numerator, but you could if you first multiplied the two 2s and two 3s in the numerator.

I cannot find a counterexample but I also cannot prove that it is always possible.

Junglemath
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  • May depend on choice of (partial) cancellation in early steps -- it seems plausible that there might always be at least one choice of ordering the cancellations so that each term selected from the denominator divides a surviving term in the numerator. – Eric Towers Aug 26 '24 at 01:59
  • To clarify; you want to avoiding splitting a factor in the denominator, so that it is canceled by two factors in the numerator. But, are you allowing a single numerator factor to cancel multiple denominator factors? In Eric's example of $\binom 84=\frac{2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8}{2\cdot 3\cdot 4\cdot 2\cdot 3\cdot 4}$, are you allowed to use the $8$ in the numerator to cancel both a $2$ and a $4$ in the denominator? – Mike Earnest Aug 26 '24 at 02:53
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    @MikeEarnest Yes, you can use a single factor to cancel two different ones in the denominator. What I want to avoid is a situation in which you need to do some multiplication in the numerator in order to cancel something in the denominator. Pretty much any sequence of cancellations in which at the end the denominator is 1 (as it must be) but without ever doing any multiplications in the numerator along the way. – Junglemath Aug 26 '24 at 03:07
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    It is possible to cancel all powers of any fixed prime $p$ - see here where I give an algorithm to rewrite a binomial coef as a product of fractions whose denominators are not divisible by $p.\ \ $ – Bill Dubuque Aug 27 '24 at 08:21

2 Answers2

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The factors cannot always be canceled as you describe.

Consider $\frac{22!}{4!9!9!}$. We must place 2 eights and nines, taking (8, 9, 16, 18) and leaving us with (2, 2). We must also place 2 sixes, and with 18 gone this takes (6, 12) leaving (2). Finally, we must place 3 fours, and we only have (4, 20) to place them, as (8, 12, 16) are unavailable.

Angelica
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  • Can this be adapted to produce counterexamples for $r > 3$? $r = 2$? – ronno Aug 26 '24 at 12:41
  • @ronno come to think of it, $\frac{16!}{8!^2}$ is impossible because of placing the (8, 6, 4). I think you could do it for larger $r$ by increasing the power of 2, and then tacking on as many 2!'s as you can. E.g. $\frac{70!}{32!^22!^3}$ must place the 32s and 24s first, then 16s are impossible. – Angelica Aug 26 '24 at 12:54
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    What does "leaving us with (2, 2)" mean? – no comment Aug 26 '24 at 17:59
  • @nocomment Per OP's comment under the question, terms in the numerator may be used to cancel multiple denominator terms, as long as you don't combine numerator terms. "Leaving (2, 2)" means that after canceling some other denominator terms, the 8, 9, 16, and 18 in the numerator are only able to cancel two 2s – Angelica Aug 26 '24 at 18:06
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Depends on the order of early cancellations. Example: \begin{align*} \frac{8!}{4!4!} &= \frac{1 \cdot 2 \cdot 3 \cdot \color{blue}4 \cdot 5 \cdot 6 \cdot 7 \cdot \color{red}8}{4! \cdot 1 \cdot \color{blue}2 \cdot 3 \cdot \color{red}4} \\ &= \frac{1 \cdot 2 \cdot 3 \cdot 2 \cdot 5 \cdot 6 \cdot 7 \cdot 2}{4! \cdot 1 \cdot 3} \end{align*} Now all the factors of $2$ are split in the numerator and there is no $4$ for the $4$ in $4!$ to cancel with.

Of course, we could have chosen a different initial cancellation to preserve the $4$ -- say combining the highlighted $2$ and $4$ into an $8$ to cancel the largest thing in the numerator. I suspect this strategy generically works -- cancel the largest factor in the numerator that you can (with however many factors, preferably using composites with many prime factors, from the denominator) so that you use up larger powers in the denominator, leaving only small powers there. Of course, this is not a proof and there may be additional subtleties.

Eric Towers
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  • If one is looking for a minimal example of order dependence, I think $7!/(4!3!)$ is it -- with the $2$ in $3!$ splitting $4$ in the numerator -- or not. – Eric Towers Aug 26 '24 at 02:41
  • I was thinking about this idea of canceling the largest possible factor in the numerator for each factor in the denominator - in your example that would mean the 4 would cancel with the 4 and not with the 8. Any idea for a proof that this would work in general? Also, if you do this, the one 4 that you highlighted in red could indeed cancel with the 8 and the 4 (in the unexpanded 4!) can be canceled with the 4 in the numerator. – Junglemath Aug 26 '24 at 02:44
  • I think a greedy algorithm will work but don't see how to prove it. The challenges are large primes and large powers of small primes, so start with those. Find the largest prime dividing any $k_i$. From Legendre's theorem there are at least as many factors in the numerator that include this prime as in the denominator. Match them up starting from the smallest factor in the numerator that will accept the one of interest from the denominator. Similarly for the highest power of each smaller prime. It seems obvious that you won't get stuck but I don't see a proof. – Ross Millikan Aug 26 '24 at 02:57
  • @RossMillikan Why is Legendre's theorem necessary here? By the combinatorial argument that the whole expression is an integer, it must necessarily be that all primes (including multiplicity) in the denominator appear at least as many times in the numerator. The question is only if those primes are condensed into a single integer or spread out over different integers in the numerator. – Junglemath Aug 26 '24 at 03:03
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    @Junglemath : That's not "the only question". The question is: Can one find an algorithm for cancellation (simpler that enumerating the method for each possible ratio of factorials) that at each step leaves enough sufficiently high powers of primes? This Answer shows that not all cancellation sequences do so. – Eric Towers Aug 26 '24 at 04:29