Combinatorics submultisets Inclusion-Exclusion

Question

Find the number of submultisets of {$25 \cdot a, 25 \cdot b, 25 \cdot c, 25 \cdot d$} of size $80$.

I applied Inclusion-Exclusion to get; $$ {80+3\choose 3} - {4\choose1}\cdot{80-26+3\choose3} + {4\choose2}\cdot{80-2\cdot26 +3\choose 3}- {4\choose3}\cdot{80 - 3\cdot26 + 3 \choose 3}$$

Is this approach and answer correct?

Answer 1

As already noted in a comment, your approach and your answer are correct. This can be viewed as a problem of balls in bins with limited capacity, where in this case all four bins have the same capacity $25$, so we can use the equation given at the end of that page:

$$ N(k)=\sum_{t=0}^m(-1)^t\binom mt\binom{m+k-t(R+1)-1}{m-1}, $$

where $k=80$ is the number of balls, $m=4$ is the number of bins and $R=25$ is the capacity of each bin, and, contrary to the usual convention, binomial coefficients with negative upper index are taken to be $0$. This yields

$$ N(80)=\binom{4+80-1}{4-1}-\binom41\binom{4+80-26-1}{4-1}+\binom42\binom{4+80-2\cdot26-1}{4-1}-\binom43\binom{4+80-3\cdot26-1}{4-1}=1771\;, $$

exactly as you calculated.

Answer 2

The solution with the inclusion-exclusion principle is correct, but a bit long. A simple solution is about finding the number of non-negative solutions of $x+y+z+t = 100-80=20$. It is $$\dbinom{23}{3} = 1771 .$$