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As a multinomial coefficient,
a1!a2!⋯ak!(a1+a2+⋯+ak)!
is an integer for any nonnegative integers a1, a2, …, ak. Taking ai=n for all i, you get that
(n!)k(nk)!
is an integer.
Let us assume that we have a1 number of objects of 1 type , a2number of objects of some type and ... ak number of objects of some other type;
So We Get Total Number Of Objects As
TotalObjects=(a1+a2+a3+....+ak)
Therefore Using Combinatorics We Get Number Of Ways To Arrange The Objects As
a1!a2!a3!....ak!(a1+a2+a3+........+ak)!
Number Of Ways To Arrange Will Be Obviously An Integer , Hence We Have Proved
That
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
As a multinomial coefficient, a1!a2!⋯ak!(a1+a2+⋯+ak)! is an integer for any nonnegative integers a1, a2, …, ak. Taking ai=n for all i, you get that (n!)k(nk)! is an integer.
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i was not aware about this theorem , so i had used combinatorics to prove the above question thanks for sharing
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Can you share your combinatorics approach?
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a1!a2!a3!....ak!(a1+a2+a3+........+ak)!
Using Combinatorics
Let us assume that we have a1 number of objects of 1 type , a2number of objects of some type and ... ak number of objects of some other type; So We Get Total Number Of Objects As
TotalObjects=(a1+a2+a3+....+ak)
Therefore Using Combinatorics We Get Number Of Ways To Arrange The Objects As
a1!a2!a3!....ak!(a1+a2+a3+........+ak)!
Number Of Ways To Arrange Will Be Obviously An Integer , Hence We Have Proved That
a1!a2!a3!....ak!(a1+a2+a3+........+ak)!
Will Always Be An Integer