341This proof holds only when p and q are relatively prime whereas a combinatorial argument will easily show that this is true for any pair (p,q)
If p and q are any two positive integers ,show that (pq)! is divisible by (p!)q.q! and by (q!)p.p!
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3 Answers
1(pq)! = (pq)(pq-1)...(1)
note the above expansion will also contain q! ~~~~~~~(X)
also the above expansion has pq terms
so the terms can be grouped into q groups with p CONSECUTIVE terms each
now you must be knowing product of k consecutive terms is a multiple of k!
so p! divides each group.
and there are q groups in all so (p!q) divides (pq)! ~~~~~~~~~~(Y)
from (X) and (y) we get
(p!q)q! divides (pq)!
the second part can be proved similarly
this proof is on the lines of prophet sir's proof here
http://targetiit.com/iit-jee-forum/posts/permutation-combination-10320.html
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