Factorial Divided By Factorial And That To So Many Times

Can You Prove That,

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

#Combinatorics #NumberTheory #Factorial #Divisibility

Note by Harsh Depal
7 years, 3 months ago

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Comments

As a multinomial coefficient, (a1+a2++ak)!a1!a2!ak!\frac{(a_1 + a_2 + \dots + a_k)!}{a_1! a_2! \dotsm a_k!} is an integer for any nonnegative integers a1a_1, a2a_2, \dots, aka_k. Taking ai=na_i = n for all ii, you get that (nk)!(n!)k\frac{(nk)!}{(n!)^k} is an integer.

Jon Haussmann - 7 years, 3 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 - 7 years, 3 months ago

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Can you share your combinatorics approach?

Calvin Lin Staff - 7 years, 2 months ago

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@Calvin Lin Yes we can prove ,

(a1+a2+a3+........+ak)!a1!a2!a3!....ak!\\ \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 a1{ a }_{ 1 } number of objects of 1 type , a2{ a }_{ 2 }number of objects of some type and ... ak{ a }_{ k } number of objects of some other type; So We Get Total Number Of Objects As

TotalObjects=(a1+a2+a3+....+ak)\\ \\ 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

(a1+a2+a3+........+ak)!a1!a2!a3!....ak!\\ \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

(a1+a2+a3+........+ak)!a1!a2!a3!....ak!\\ \frac { ({ a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+........+{ a }_{ k })! }{ { a }_{ 1 }!{ a }_{ 2 }!{ a }_{ 3 }!....{ a }_{ k }! }

Will Always Be An Integer

Harsh Depal - 7 years, 2 months ago
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