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

$\text{7 digits, digit sum } = 10$ $\text{allowed digits } : 1, 2, 3$ $\text{Count all such 7-digit numbers}$ $\text{options } : 55,\ 66,\ 77,\ 88$ $a \text{ ones},\quad b \text{ twos},\quad c \text{ threes}$ $a+b+c = 7$ $a\cdot 1 + b\cdot 2 + c\cdot 3 = 10$ $a+2b+3c = 10$ $(a+2b+3c)-(a+b+c) = 10-7$ $b+2c = 3$ $\text{no } 0 \text{ used } \Rightarrow \text{every arrangement is a valid 7-digit number}$ $c=0:\ a=4,\ b=3$ $c=1:\ a=5,\ b=1$ $\text{arrangements} = \dfrac{7!}{(\text{repeats})!}$ $\dfrac{7!}{4!\,3!} = 35$ $\dfrac{7!}{5!\,1!\,1!} = 42$ $35 + 42 = 77$ $\text{Answer: option (3) } 77$