Answer: (1)

Solution

Case I: Six 1’s and one 3: $\frac{7!}{6!} = 7$ arrangements. Wait – sum = 10 with 7 digits using 1,2,3. Case I: $1,1,1,1,1,2,3$ (sum = 10): ways $= \frac{7!}{5!} = 42$. Case II: $1,1,1,1,1,1,2,2$ – only 7 digits so $1,1,1,1,1,2,2$ (sum = 9, no). Recount: digits sum to 10 with 7 digits from {1,2,3}. If all 1’s: sum=7. Need 3 more distributed. $(0,0,3) \to$ one digit becomes $1+3=4$ – no. Use substitution: replace 1 by 2 adds 1, replace 1 by 3 adds 2. Need additions = 3. Cases: (i) three 2’s rest 1’s: $1,1,1,1,2,2,2$, ways $= \frac{7!}{4!3!} = 35$. (ii) one 2 and one 3 rest 1’s: $1,1,1,1,1,2,3$, ways $= \frac{7!}{5!} = 42$. Total $= 35 + 42 = 77$.