Total number of solutions for a given equation with a minimum value $0$ and a maximum value of $m$

Question

What is the total number of solutions for following equation

$$x_1+x_2+x_3+.....+x_k=n$$

with a maximum value of $m$, where $0\leq x \leq m$. No two variables should contain a maximum value $m$.

Example: $x_1+x_2+x_3=3$ with a maximum value $2$ ($n=3$, $m=2$, $k=3$)

Total possible ways is $7$: $$\{1,1,1\}, \{0,1,2\}, \{0,2,1\}, \{1,0,2\}, \{1,2,0\}, \{2,0,1\}, \{2,1,0\}$$

I've done it manually but I don't how to come up with a formula. I've been struggling for last one week. Is there any formula?

Sorry I don't know about generating functions.Small correction is we can have x1+x2+...+xk variables not limited to m varables — Malka, Mar 19 '13 at 14:15
@Malka: You can edit the question with the correction. I have done it for you. — ShreevatsaR, Mar 19 '13 at 14:17
Why is ${0, 1, 2}$ a valid solution, but ${0, 1, 1, 1}$ is not? I think your problem definition is missing something. — Jonathan Rich, Mar 19 '13 at 14:19
@JonathanRich:We have only 3 variables x1,x2 and x3 right and total sum we need is 3 — Malka, Mar 19 '13 at 14:23
@ShreevatsaR:Thank you,actually this is the first time I'm posting a question on any blog — Malka, Mar 19 '13 at 14:24
So $k$ is also fixed (you have three parameters: $n$,$m$, $k$)? — leonbloy, Mar 19 '13 at 14:25
@leonbloy:yes k is also fixed.I've n,m and k values and need total number of solutions — Malka, Mar 19 '13 at 14:36

score 3 · Answer 1 · edited Apr 13 '17 at 12:20

3

Hint: You can divide the solutions into two groups. One, those which attain the maximum value ($x_i = m$ for some $i$, and $x_j < m$ for all $j\ne i$); the other, in which $x_i < m$ for all $i$.

Then you can count the cardinality of each set by looking into this question.

Specifically, adapting the linked pdf (formula E, page 441), we get that the number of compositions of $k$ numbers $0\le x_i<m$ with $\sum x_i = n$ is given by

$$F(n,m,k)= \sum_{j=0}^{\lfloor n/m \rfloor} (-1)^j {k \choose j}{n+k - j \, m -1 \choose k-1}$$

Then we have

$$C(n,m,k) = k \, F(n-m,m,k-1) + F(n,m,k) $$

In your example:

$$C(3,2,3) = 3 \, F(1,2,2) + F(3,2,3) = 3 \times 2 + 1 = 7$$

edited Apr 13 '17 at 12:20

Community

1

answered Mar 19 '13 at 14:47

leonbloy

66,202

:Will this formulat considers condition "Only one variable out of k variables contains maximum value m"? – Malka Mar 21 '13 at 06:15
Example:x1+x2+x3+x4=6 with max value 3.n=6,k=4,m=3.Solutions {3,3,0,0},{3,0,3,0},{0,3,3,0},{0,0,3,3} are invalid – Malka Mar 21 '13 at 06:16
Yes, that restriction is taken into account, see the hint, the two groups correspond (respectively) to the two terms in $C(n,m,k)$ formula – leonbloy Mar 21 '13 at 11:14
... and, regarding your first comment, I gather that it actually should read "at most one variable" instead of "only one variable". – leonbloy Mar 21 '13 at 11:45
Is there any generic formula for given both minimum value and maximum values. – Malka Apr 30 '13 at 08:52
What about marking you past questions as answered before asking new questions? http://math.stackexchange.com/faq#howtoask – leonbloy Apr 30 '13 at 13:24

Total number of solutions for a given equation with a minimum value $0$ and a maximum value of $m$

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