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In fact, I couldn't prove the inequality because I don't know which method is used for this. The condition for this inequality is $$(a+b+c)>1$$

E.H.E
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7 Answers7

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$$\frac{(a+b+c)!}{a!b!c!} $$ is the number of ways to arrange $a$ (indistinguishable) amber balls, $b$ blue balls and $c$ cyan balls in a line. Your goal is to show that this number is $>1$. This is the case as soon as at least two of $a,b,c$ are positive.

Hagen von Eitzen
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  • Ok, but you proved the inequality by itself, this is impossible. – E.H.E Oct 30 '14 at 17:34
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    @Ehegh: What precisely is your objection against the argument? Hagen realized the number in question as the cardinality of a set which by direct inspection can be seen to have more than one element. For me that counts as a proof. – Hanno Oct 30 '14 at 17:49
  • I mean that the $$\frac{(a+b+c)!}{a!b!c!}>1$$ need a proving – E.H.E Oct 30 '14 at 17:52
  • @Ehegh: Hagen is arguing that 1) the inequality is equivalent to a certain quantity being larger than one, 2) this quantity reflects the correct counting for a certain set of objects, 3) this set of objects contains more than one object if at least two of $a,b,c$ are positive integers. If you think at least one of these steps isn't justified, you should specify which one. – Semiclassical Oct 30 '14 at 19:59
  • My question was permutation and the proving of Hagen also was permutation. If you prove the $\frac{(a+b+c)!}{a!b!c!}>1$ is right the proving will become right.I mean that the Hagen inequality needs proving. – E.H.E Oct 30 '14 at 20:09
  • @Ehegh Indeed. He is inviting you to prove that yourself since it is quite easy to do so. If you have the balls in question, can you show that the number of ways to order the balls will be more than one? All you have to do is show that there are at least two ways to order them to prove the statement. – MT_ Oct 31 '14 at 02:32
  • @Hanno: For me, that even counts as a nice proof. :) – anomaly Nov 20 '15 at 18:26
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This will help:

$$(a+b)! = a!(a+1)(a+2)...(a+b)<a!(1)(2)...(b)=a!b!.$$

Your turn to prove it for $(a+b+c)!$...

$$(a+b+c)! = a![(a+1)(a+2)...(a+b)][((a+b)+1)((a+b)+2)((a+b)+c)].$$

Vladimir Vargas
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Show that $(x+y)! > x!y!$:

Divide by $x!$ to get $(x+1)(x+2)(x+3)\dots(x+y) > y! = (1)(2)(3)\dots(y)$, which is true for $x > 0$ (compare terms pairwise!)

Let $x = a, y = b+c$. Then we have $(a+b+c)! > a!(b+c)!$

Furthermore we have $(b+c)! > b! c!$, so $(a+b+c)! > a!(b+c)! > a!b!c!$.

MT_
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$(a+b+c)!$ and $a!b!c!$ are both products of $a+b+c$ terms. Can you see a way of comparing them individually?

preferred_anon
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Hint: $$(a+b+c)!= a! \prod_{k=a+1}^{a+b} k! \prod_{j=a+b+1}^{a+b+c}j!, $$ where $\prod_{k=a+1}^{a+b}$ is products of $b$ terms each of which is at least $b$, hence $\prod_{k=a+1}^{a+b}\geq b^b>b!$...

Similarly for $\prod_{j=a+b+1}^{a+b+c}j!>c!$.

Milly
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Use induction to show that $(a+b)! > a!b!$. You'll need to induce over both a and b. Note that this is for $a \ge 1, b \ge 1$.

Daniel Goldman
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An adaptation of this answer. If you have $a$ adults, $b$ boys, and $c$ girls, then $(a+b+c)!$ counts the number of ways you can line everyone up, while $a!b!c!$ counts the number of ways you can do so with the additional requirement that all adults come before all children, and all boys before all girls. Then $(a+b+c)!\geq a!b!c!$ always holds, while $(a+b+c)!>a!b!c!$ holds as soon as at least two categories are non-empty, since then there will be some line-up that violates the additional requirement.

Community
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Marc van Leeuwen
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