The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only is:
Solutions · 1
Official solution
7 digits from {1,2,3} summing to 10 (minimum sum with all 1's is 7, so 3 extra needed).
Case 1: digits 1,1,1,1,1,2,3 (four extra 1's stay, one becomes 2, one becomes 3 --- sum =5+2+3=10): arrangements =5!7!=42
Case 2: digits 2,2,2,1,1,1,1 (three 2's, four 1's; sum=6+4=10): arrangements =3!4!7!=35
\[\text{Total} = 42+35 = 77\]