number of integer solutions combinatorial problem

Question

Find the number of integer solutions to $x_1+x_2+x_3+...+x_7=23$ subject to $x_1\gt0,x_2\ge3$ and $x_i\gt0$ for all $i\ge3$.

This is the given answer in the book:
Using the substitutions $y_1=x_1-1$ and $y_2=x_2-3$ problem is equivalent to solving $y_1+y_2+..+y_7=19$ subject to $y_i\ge0$ for all $i$. Therefore number of solutions are ${19+6-1 \choose 6-1}={24 \choose 5}$.

But this is my answer. I made the substitutions $y_2=x_2-3$ and $y_i=x_i-1$ for $i={1,3,4,5,6,7}$.
Then the original problem is equivalent to solving $y_1+y_2+y_3+y_4+y_5+y_6+y_7+9=23$ subject to $y_i\ge0$ for all $i$..Therefore number of solutions are ${14+7-1 \choose 7-1}={20 \choose 6}$.

I don't understand the answer in the book. Is my answer wrong? If the substitution for $x_1\gt0$ is made then why isn't it done for $x_i\gt0$ for $i\ge3$.

Also why is $6$ partitions considered there in ${19+6-1 \choose 6-1}$ and not $7$?
Shouldn't it be $7$?

score 1 · Answer 1 · answered Oct 11 '14 at 15:03

1

If the question you wrote is correct, I think the given answer is wrong.

Let $X_i=x_i-1$ for $i\not=2$ and $X_2=x_2-3$. Then, $X_i$ are non-negative integers and we have $$\sum_{i=1}^{7}X_i=23-9=14.$$ Hence, the answer is $$\binom{7+14-1}{14}=\binom{20}{6}=38760.$$

answered Oct 11 '14 at 15:03

mathlove

139,939

I too should get this answer.I've made a mistake.It is necessary that $x_i>=0$ and not $x_i>0$ right? – clarkson Oct 11 '14 at 15:10
@clarkson: I think even if $x_1\gt 0,x_2\ge 3,x_i\color{red}{\ge} 0\ (i\ge 3)$ in the question, the given answer is wrong. – mathlove Oct 11 '14 at 15:28
:Yes it is still wrong – clarkson Oct 11 '14 at 15:34

number of integer solutions combinatorial problem

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