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Factorial Divided By Factorial And That To So Many Times

Can You Prove That,

\((nk)!\) is always divisible by \({ (n!) }^{ k }\)

Note by Harsh Depal
3 years, 10 months ago

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As a multinomial coefficient, \[\frac{(a_1 + a_2 + \dots + a_k)!}{a_1! a_2! \dotsm a_k!}\] is an integer for any nonnegative integers \(a_1\), \(a_2\), \(\dots\), \(a_k\). Taking \(a_i = n\) for all \(i\), you get that \[\frac{(nk)!}{(n!)^k}\] is an integer.

Jon Haussmann - 3 years, 10 months ago

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i was not aware about this theorem , so i had used combinatorics to prove the above question thanks for sharing

Harsh Depal - 3 years, 10 months ago

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Can you share your combinatorics approach?

Calvin Lin Staff - 3 years, 10 months ago

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@Calvin Lin Yes we can prove ,

\(\\ \frac { ({ a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+........+{ a }_{ k })! }{ { a }_{ 1 }!{ a }_{ 2 }!{ a }_{ 3 }!....{ a }_{ k }! } \)

Using Combinatorics

Let us assume that we have \({ a }_{ 1 }\) number of objects of 1 type , \({ a }_{ 2 }\)number of objects of some type and ... \({ a }_{ k }\) number of objects of some other type; So We Get Total Number Of Objects As

\(\\ \\ Total\quad Objects\quad =\quad ({ a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+....+{ a }_{ k })\)

Therefore Using Combinatorics We Get Number Of Ways To Arrange The Objects As

\(\\ \frac { ({ a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+........+{ a }_{ k })! }{ { a }_{ 1 }!{ a }_{ 2 }!{ a }_{ 3 }!....{ a }_{ k }! } \)

Number Of Ways To Arrange Will Be Obviously An Integer , Hence We Have Proved That

\(\\ \frac { ({ a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+........+{ a }_{ k })! }{ { a }_{ 1 }!{ a }_{ 2 }!{ a }_{ 3 }!....{ a }_{ k }! } \)

Will Always Be An Integer

Harsh Depal - 3 years, 10 months ago

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