Thank you for reporting, we will resolve it shortly
Q.
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
Permutations and Combinations
Solution:
Solution: Let $x_{ i }(1 \leq i \leq 7)$ be the digit at the $i$ th place. As $1 \leq x_i \leq 3$, and $x_1+x_2+\ldots+x_7=10$, at most one $x_i$ can be 3 .
Two cases arise.
Case 1.
Exactly one of $x_i$ 's is 3 . In this case exactly one of the remaining $x_i$ 's is 2 .
In this case, the number of seven digit numbers is
$\frac{7 !}{5 !}=7 \times 6=42$
Case 2.
None of $x_i$ 's is 3 .
In this case exactly three of $x_i$ 's is 2 and the remaining four $x_i$ 's are 1.
In this case, the number of seven digit number is
$\frac{7 !}{3 ! 4 !}=35$
Hence, the required seven digit numbers is $42+35=77$.