Let us have \(2014!\) balls. Let us divide them in \(2012!\) groups, each group having \(2013.2014\) balls.

Let's assume that balls of each group are of different colour, but all balls of the same group are the of the same colour.

Number of ways of doing so is, \[\dfrac{(2014!)!}{((2014.2013)!)^{2012!}}\]
Number of ways of doing something is always a positive integer. Hence Proved. \(\boxed{.}\)

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TopNewestLet us have \(2014!\) balls. Let us divide them in \(2012!\) groups, each group having \(2013.2014\) balls.

Let's assume that balls of each group are of different colour, but all balls of the same group are the of the same colour.

Number of ways of doing so is, \[\dfrac{(2014!)!}{((2014.2013)!)^{2012!}}\] Number of ways of doing something is always a positive integer. Hence Proved. \(\boxed{.}\)

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Using Combinatorics for a Number Theory Problem is simply amazing!

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Excellent! +1

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@Satvik Golechha this is great method see solution of @John Ashley Capellan here

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superb way to do this......

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