Number Theory Question Possibly related to combinatorics too.

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SUMMARY

The discussion centers on proving that a! b! divides (a+b)!. The user approaches the problem by considering the factorial definitions and the relationship between the terms involved. They establish that (a+b)! can be expressed as the product of consecutive integers from 1 to (a+b), and through this, they demonstrate that the division by a! yields a product of consecutive integers starting from (a+1) to (a+b). This confirms that b! divides the resulting product, thus proving the original assertion.

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Homework Statement



Prove that [itex]a! b! | (a+b)![/itex].


Homework Equations



Probably some Number Theory Theorem I can't think of at the moment.


The Attempt at a Solution



Without loss of generality, let [itex]a < b[/itex].
Therefore [itex]b! | \Pi _{k=1}^b a+k[/itex]. Since [itex](a+1) ... (a+b)[/itex] are [itex]b[/itex] consecutive terms, [itex]b|\Pi _{k=1}^b a+k[/itex]. The problem I am running into is that you don't necessarily get [itex]b-1[/itex] consecutive terms in the next step of the recursive reasoning. Any suggestions? Thanks!

This question reminds me of questions like, how many possible combinations can you rearrange the letters in "MISSISSIPPI"? This might be a combinatorics question too but I'm not entirely sure.
 
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Consider that (a+b)! = (1)(2)(3)...(a)(a+1)(a+2)...(a+b).

And a! = (1)(2)(3)...(a).

Now it should be clear that (a+b)!/a! = (a+1)(a+2)...(a+b). Then your assertion that b! divides that should be sufficient.
 

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