Ten identical balls are placed in 4 different boxes so that no box holds more than 4 balls (a box may be empty). In how many ways can this be done?

Without the cap: ${}^{13}{C}_3=286$. A box with 5 or more: give it 5 first, then 5 balls freely, $4\times {}^8{C}_3=224$. Two boxes with 5 or more: both get 5, nothing left, ${}^4{C}_2\times 1=6$. Inclusion–exclusion: $286-224+6=68$. (62 forgets to add back the 6 doubly-removed cases.)