×

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 ago

Comments

Sort by:

Top Newest

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. · 3 years ago

Log in to reply

i was not aware about this theorem , so i had used combinatorics to prove the above question thanks for sharing · 3 years ago

Log in to reply

Can you share your combinatorics approach? Staff · 3 years ago

Log in to reply

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 · 3 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...