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. \(\boxed{.}\)
–
Satvik Golechha
·
1 year, 10 months ago

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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. \(\boxed{.}\) – Satvik Golechha · 1 year, 10 months ago

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– Marc Vince Casimiro · 1 year, 10 months ago

Using Combinatorics for a Number Theory Problem is simply amazing!Log in to reply

– Pratik Shastri · 1 year, 10 months ago

Excellent! +1Log in to reply

@Satvik Golechha this is great method see solution of @John Ashley Capellan here – Shivamani Patil · 1 year, 10 months ago

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– Rajat Bhagat · 1 year, 9 months ago

superb way to do this......Log in to reply