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 ago

Log in to reply

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

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

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 ago

## Comments

Sort by:

TopNewestAs 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 ago

Log in to reply

– Harsh Depal · 3 years ago

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

– Calvin Lin Staff · 3 years ago

Can you share your combinatorics approach?Log in to reply

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

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 ago

Log in to reply