Mathematics - Permutation & Combination Question with Solution | TestHub

There are two possible cases arise: Case I Five 1's, one 2's, one 3's $\therefore$ Number of numbers $=\frac{7!}{5!}=42$ Case II Four 1's, three 2's $\therefore$ Number of numbers $=\frac{7!}{4!3!}=35$ $\therefore$ Total number of numbers $=42+35=77$