# Operations with factorials

• June 5th 2013, 10:49 AM
jones123
Operations with factorials
Hi, i'm having troubles understanding operations with faculty.

If:

(3k)! (k+1)! (2(k+1))!
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k! (2k)! (3(k+1))!

Then why is this equal to:

(k+1)(2k+1)(2k+2)
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(3k+1)(3k+2)(3k+3)

Thanks for the help!
• June 5th 2013, 11:04 AM
Kmath
Re: Operations with factorials
Note that
$(3(k+1))!=(3k+3)!=(3k+3)(3k+2)(3k+1)(3k)!$.
And do it again with 2 instead of 3.
• June 5th 2013, 02:11 PM
Soroban
Re: Operations with factorials
Hello, jones123!

I must assume that you know how to "cancel" factorials.

Quote:

$\text{Show that: }\: \frac{\big(3k\big)!\,\big(k+1\big)!\,\big(2[k+1]\big)!}{k!\,\big(2k\big)!\,\big(3[k+1]\big)!} \;=\;\frac{(k+1)(2k+1)(2k+2)}{(3k+1)(3k+2)(3k+3)}$

$\text{We have: }\:\frac{(3k)!\,(k+1)!\,(2k+2)!}{k!\,(2k)!\,(3k+3) !}$

. . . . . $=\;\frac{(3k)!}{(3k+3)!}\cdot \frac{(k+1)!}{k!}\cdot\frac{(2k+2)!}{(2k)!}$

. . . . . $=\;\frac{1}{(3k+3)(3k+2)(3k+1)}\cdot\frac{k+1}{1} \cdot\frac{(2k+2)(2k+1)}{1}$

. . . . . $=\;\frac{(k+1)(2k+1)(2k+2)}{(3k+1)(3k+2)(3k+3)}$

But this reduces further . . .

. . . . . $=\;\frac{(k+1)\,(2k+1)\,2(k+1)}{(3k+1)\,(3k+2)\,3( k+1)}$

. . . . . $=\;\frac{2(k+1)(2k+1)}{3(3k+1)(3k+2)}$
• June 6th 2013, 07:39 AM
jones123
Re: Operations with factorials
Thanks! :)