# 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
4 years, 3 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.

- 4 years, 3 months ago

- 4 years, 3 months ago

Can you share your combinatorics approach?

Staff - 4 years, 3 months ago

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

- 4 years, 3 months ago