Q.Prove that the product of r consecutive positive integer is divisible by r! .
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The product of some rr consecutive integers can be represented as
(n+r)(n+r−1)⋯(n+1)r consective integers=(n+r)!n!(n+r)(n+r−1)⋯(n+1)⏞r consective integers=(n+r)!n!
where nn is the number one less than the smallest of the consecutive integers. Now, if it is true that primes in (n+r)!(n+r)!appear just as frequently or more as in n!r!n!r!, then you are saying that for some integer kk (likely big) that (n+r)!=k⋅n!r!(n+r)!=k⋅n!r!. So your product of nn consecutive integers is
(n+r)!n!=k⋅n!r!n!=k⋅r!(n+r)!n!=k⋅n!r!n!=k⋅r!
and is therefore divisible by r!r!.
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(n+r)(n+r−1)⋯(n+1)r consective integers=(n+r)!n!(n+r)(n+r−1)⋯(n+1)⏞r consective integers=(n+r)!n!
where nn is the number one less than the smallest of the consecutive integers. Now, if it is true that primes in (n+r)!(n+r)!appear just as frequently or more as in n!r!n!r!, then you are saying that for some integer kk (likely big) that (n+r)!=k⋅n!r!(n+r)!=k⋅n!r!. So your product of nn consecutive integers is
(n+r)!n!=k⋅n!r!n!=k⋅r!(n+r)!n!=k⋅n!r!n!=k⋅r!
and is therefore divisible by r!r!.
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