#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

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,

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

Now you can win by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

jitender

Rahul Kumar
131 Points
9 years ago

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