# How many ways can the number 51 be witten as the sum of 11 positive,odd number?

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• Apr 10th 2009, 01:03 PM
gravity2910
How many ways can the number 51 be witten as the sum of 11 positive,odd number?
Hi,
Can anyone help me to find out How many ways can the number 51 be witten as the sum of 11 positive,odd numbers?

Assume that the orders make the differences.
Thanks alot.
• Apr 10th 2009, 01:28 PM
Plato
Quote:

Originally Posted by gravity2910
Can anyone help me to find out How many ways can the number 51 be witten as the sum of 11 positive,odd numbers?
Assume that the orders make the differences.

Think of the eleven variables as different cells. To make them positive integers, think of each already containing a one. That leaves forty ones to put into the eleven cells. We think of them as two ones together. By adding two ones at a time the cell remains odd.
So \$\displaystyle \binom{20+11-1}{20}\$.
• Apr 10th 2009, 01:57 PM
gravity2910
Quote:

Originally Posted by Plato
Think of the eleven variables as different cells. To make them positive integers, think of each already containing a one. That leaves forty ones to put into the eleven cells. We think of them as two ones together. By adding two ones at a time the cell remains odd.
So \$\displaystyle \binom{20+11-1}{20}\$.

Thank you very much, plato....
But I'm still not so clear. If you put 20 pairs of one into 11 positions, It's look like the way to write 20 as the sum of 11 non-negative numbers. So are the total possibilities : (20+11-1|11-1) ???????????????? So why do you have (20+11-1|20)??
• Apr 10th 2009, 02:01 PM
Plato
They equal.
\$\displaystyle \binom{20+11-1}{20}=\binom{20+11-1}{11-1}\$
\$\displaystyle \binom{N}{k}=\binom{N}{N-k}\$
• Apr 10th 2009, 02:03 PM
gravity2910
Oh, I'm so stupid....I didn't think about it.......thanks alot.