How many solutions does this equation x1+x2+x3 = 11, given x1,x2,x3 >0 have?

This is solved using combination.

How can this be solved.

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- Nov 14th 2010, 01:28 AMkumaran5555NUmber of solutions
How many solutions does this equation x1+x2+x3 = 11, given x1,x2,x3 >0 have?

This is solved using combination.

How can this be solved. - Nov 14th 2010, 01:46 AMemakarov
See Theorem one in this Wikipedia article.

- Nov 14th 2010, 02:08 AMkumaran5555
Thanks.

But it is not matching with the answer given.

By using that theorem, it would taken two from 10 available gaps, 10C2.

But i have the answer as 13c11.

I think they are considering it as X1,X2,X3 >=0. So we can have 13 gaps and 13C2 which is equal to 13C11.

I dont know whether my understanding is wrong , please correct. - Nov 14th 2010, 02:26 AMemakarov
You may be right. However, if in fact there no mistake in the problem statement, could you post an update here when you find out what's going on?

- Nov 14th 2010, 02:30 AMkumaran5555
Thanks a lot. And sure i'll update if find anything wrong about the question.

- Nov 14th 2010, 04:52 AMSoroban
Hello, kumaran5555!

Quote:

How many solutions does this equation have: . $\displaystyle x_1+x_2+x_3\:=\:11$

. . given $\displaystyle x_1,x_2,x_3 \,>\,0$

Consider an eleven-inch board marked in one-inch intervals.

. . . . $\displaystyle \square\!\square\!\square\!\square\!\square\!\squa re\!\square\!\square\!\square\!\square\!\square $

It can be divided into three nonzero pieces

. . by choosing any two of the ten inch-marks

. . and cutting the board there.

Therefore, there are: .$\displaystyle \displaystyle _{10}C_2 \:=\:{10\choose2} \:=\:\frac{10!}{2!\,8!} \:=\:45\text{ solutions.}$