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.
See Theorem one in this Wikipedia article.
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.
Hello, kumaran5555!
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: .$\displaystyle \displaystyle _{10}C_2 \:=\:{10\choose2} \:=\:\frac{10!}{2!\,8!} \:=\:45\text{ solutions.}$