Compute Card$(\{(m_1,m_2,m_3)\in\{0,\dots,75\}^3: m_1+m_2+m_3=75\})$. This question arose from this one. Now I just want to know how can I compute the cardinal of $$\{(m_1,m_2,m_3)\in\{0,\dots,75\}^3: m_1+m_2+m_3=75\}$$ (curiosity).
Any ideas?
 A: That's the number of weak compositions of $n=75$ with $k=3$ parts. General solution is $\binom{n+k-1}{k-1}$, which is $2926$ in your case, I think.
A: If you choose $m_1,m_2 \in \{0,...,75\}$ with $m_1 + m_2 \leq 75$ you are just fixing $m_3$. For each fixed choice of $m_1$ you have $m_2 \in \{0,...,75-m_2\}$. So the cardinal of our set is $\sum_{m_1=0}^{75}(76-m_1)$.
A: Imagine as if you had $3$ numbered boxes, and $75$ identical (not distinguishable) marble you want to distribute in them. Then the number of ways you can distribute them corresponds to the cardinality you seek. Imagine it with $3$ boxes and 6 marbles. For instance:
$$[XX] \, [XXX] \, [X]$$
This can also be represented this way:
$$XX|XXX|X \,\, (*)$$
And the number of permutations of permutations of $(*)$ is your answer. Which is the permutations of the number of total symbols of $(*)$ (in your case $77$ symbols), so $77!$, divided by the sub-permutations of $X$'s, which is $75!$ and of the lines $2!$ for you is $\frac{77!}{2!(75)!}=\binom{77}{2}$.
In general, for $n$ marbles and $k$ boxes we'd get, if you follow the same procedure: $\binom{n+k-1}{k-1}$. As you see, in your case, $n = 75,k=3 \Rightarrow \binom{n+k-1}{k-1} = \binom{77}{2}$.
