The number of ways of distributing 10 identical balls among 4 distinct boxes so that every box receives at least 1 ball and no box receives more than…

Let box $i$ get $x_i$ balls with $1\le x_i\le4$ and $\sum x_i=10$. Put $y_i=x_i-1$, so $y_i\ge0$, $y_i\le3$ and $\sum y_i=6$. Without the upper bound the count is $\binom{6+3}{3}=84$. Violations: some $y_i\ge4$. Fix one box with $y_i\ge4$; set $y_i'=y_i-4$, giving $\sum=2$ and $\binom{2+3}{3}=10$ solutions, and there are $4$ choices of box: $40$. Two boxes with $y\ge4$ would need total $\ge8>6$, impossible, so no overlap. By inclusion-exclusion the answer is $84-40=44$. Direct enumeration of all quadruples in $\{1,2,3,4\}^4$ with sum $10$ also gives $44$. Option (B) $84$ ignores the upper bound; (C) $40$ is the number of violating arrangements; (D) arises from wrongly using $\binom{8}{3}$.