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How many integer solutions to the equation $x_1 + x_2 + x_3 = 15$ so that $x_1 \ge 3, x_2 \ge 2$ and $x_3 \ge 0$ ? I honestly don't know where to start on this

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up vote 4 down vote accepted

HINT: Let $y_1=x_1-3$, $y_2=x_2-2$, and $y_3=x_3$. Note that $x_1+x_2+x_3=15$ if and only if $y_1+y_2+y_3=10$, and integers $x_1,x_2$, and $x_3$ satisfy the inequalities $x_1\ge 3$, $x_2\ge 2$, and $x_3\ge 0$ if and only if the corresponding $y_1,y_2$ and $y_3$ are non-negative. Now use the standard stars-and-bars solution to the problem of counting solutions to $y_1+y_2+y_3=10$ in non-negative integers.

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okay, so does 10C2 sound about right – Cintdy T Mar 2 at 3:28
@CintdyT: Not quite: it should be $\dbinom{10+3-1}{3-1}=\dbinom{12}2$. – Brian M. Scott Mar 2 at 5:48

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