Let $P$ be the set of seven-digit numbers with sum of their digits equal to $11$. If the numbers in $P$ are formed by using the digits $1,2$ and $3$ only, then the number of elements in $P$ is:
Step-by-Step Solution
Key Concept: Let $x_{i}=$ count of digit $i$. System: $x_{1}+x_{2}+x_{3}=7$ and $x_{1}+2x_{2}+3x_{3}=11.$ Subtract to get $x_{2}+2x_{3}=4$; enumerate $x_{3}\in\{0,1,2\}.$
Subtract: $x_{2}+2x_{3}=4.$
$x_{3}=0,\,x_{2}=4,\,x_{1}=3$: arrangements $\dfrac{7!}{3!4!}=35.$
$x_{3}=1,\,x_{2}=2,\,x_{1}=4$: arrangements $\dfrac{7!}{4!2!1!}=105.$
$x_{3}=2,\,x_{2}=0,\,x_{1}=5$: arrangements $\dfrac{7!}{5!0!2!}=21.$
Total: $35+105+21=161.$
Correct Answer: 4