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Suppose we have the equation $$ x + y + z + u = 29 $$ subject to the constraints $$ x \geq 1 \;,\quad y \geq 2 \;,\quad z \geq 3 \; \quad\text{and}\quad u \geq 0 \;. $$ How do we find the number of solutions that this equation has? A step by step explanation of the concepts will be appreciated as I am new to Linear Algebra. Thanks.

Björn Friedrich
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sumitpal0593
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    In $\mathbb Z$ ? –  Sep 15 '18 at 11:43
  • That is not mentioned in the question. – sumitpal0593 Sep 15 '18 at 11:51
  • Then there is a vast infinity of solutions. –  Sep 15 '18 at 12:04
  • That's what I thought as well but there are four options associated with the question: a)4960 b)2600 c)23751 and d)8855 – sumitpal0593 Sep 15 '18 at 12:06
  • This is a number theory question then. –  Sep 15 '18 at 12:45
  • And it's a question of a type thaat has been asked and answered on this site many times before. See, for example, https://math.stackexchange.com/questions/2274456/find-the-number-of-solution-of-equation and the questions linked there. – Gerry Myerson Sep 15 '18 at 12:58
  • Hint: $23751=\frac{26\cdot27\cdot28\cdot29}{1\cdot2\cdot3\cdot4}$. –  Sep 15 '18 at 13:00
  • @YvesDaoust that is a rather misleading hint. – JMoravitz Sep 15 '18 at 13:42
  • As for the topic of there being infinitely many solutions, there would be infinitely many solutions if we didn't have the limitation that $x,y,z,u$ were strictly non-negative integers. If one or more were allowed to be negative integers, or if we were taking solutions over the rationals or reals for example, then we would of course have infinitely many solutions. That there are only finitely many solutions for the actual problem being asked should be obvious as $30^4$ is an easy finite upper bound to the number of solutions as there are only $30^4$ tuples of nonnegative integers less than 30 – JMoravitz Sep 15 '18 at 13:46

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Change the problem from finding integer solutions of $\begin{cases} x+y+z+u=29\\x\geq 1\\y\geq 2\\z\geq 3\\u\geq 0\end{cases}$

to instead finding the number of integer solutions to $\begin{cases}x'+y'+z'+u'=23\\x'\geq 0\\y'\geq 0\\z'\geq 0\\u'\geq 0\end{cases}$

From here, apply the technique of stars-and-bars.

With $k$ unknowns adding up to $n$ where each unknown may be any non-negative integer, there are $\binom{n+k-1}{k-1}$ solutions. In our specific case, that is $\binom{23+4-1}{4-1}=\binom{26}{3}=2600$ solutions.

JMoravitz
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