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PROVE THAT (3n)! is divisible by n!*(n+1)!*(n+2)!.

PROVE THAT (3n)! is divisible by n!*(n+1)!*(n+2)!.


3 Answers

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

Hi Aritra,


This is not true.

Check that for n=1, n=2....... (3n)! will not be divisible by n!(n+1)!(n+2)!....


This result holds true only for n ≥ 3, and is quite easy to prove.

Use that the product of n-consecutive numbers, is always divisible by n!.


Hope this helps.


Best Regards,

Ashwin (IIT Madras).

jitender lakhanpal
62 Points
9 years ago

Dear aritra,

(3n)! = (3n)(3n-1)(3n-2)(3n-3)(3n-4)(3n-5)(3n-6)...............(2n)(2n-1)(2n-2)(2n-2)(2n-3).......(n+2)(n+1)n!----1

take the terms 3n,3n-3,3n-6,3n-9.....upto 3   and 2n,2n-2,2n-4.........upto 2 

together   and take 3 common from each   and similarly 2 common from each

so we get  n! twice and there is one more n! is already present so when we divide n!*(n+1)!*(n+2)!

all the three n! gets cancelled and remaining n+1, n+2 is also present in the equation 1

so n!*(n+1)!*(n+2)! divides 3n! completely  ------hence proved

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Rahul Kumar
131 Points
9 years ago

The problem is wrong.Take an example and verify.It us nt verified.

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