Prove that it is integer

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Prove that (a₁+a₂+...+a_n)!a₁!a₂!...a_n! is an integer

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Hardik Sheth ·2013-02-23 08:56:41

this is the form for the no of arrangements of a1+a2...a_n things wer a1 are of one type,a2 are of other type and so on...hence expression is an integer

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Shaswata Roy ·2013-02-22 23:51:39

We know that the product of r consecutive integers are divisible by r!.
The numerator consists of a₁+a₂+....+a_n consecutive integers.

The first set of a₁ consecutive integers are divisible by a₁!.
The second set of a₂ consecutive integers are divisible by a₂!.
...
The n^th set of a_n consecutive integers are divisible by a_n!.

Hence the numerator is divisible by a₁!*a₂!*a₃!*......*a_n!

Hence the given number is an integer.

It can also be said that the given number is the coefficient of x₁^a1*x₂^a2.....x_n^an of the expansion of :

(x₁+x₂+.....x_n)^{a1+a2+a3...+an} which is always an integer.

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Basic Integral Calculus Operators Symbols Matrix Cases

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Prove that it is integer

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