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
·
2 years, 1 month 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 · 2 years, 1 month ago

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– Marc Vince Casimiro · 2 years, 1 month ago

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

– Pratik Shastri · 2 years, 1 month ago

Excellent! +1Log in to reply

@Satvik Golechha this is great method see solution of @John Ashley Capellan here – Shivamani Patil · 2 years, 1 month ago

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– Rajat Bhagat · 2 years ago

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