Let $P$ be the set of all 7-digit numbers that can be formed using only the digits 1, 2, 3 with digit sum equal to 11. Then $|P|$ equals
Step-by-Step Solution
Key Concept: Let $n_1, n_2, n_3$ be the counts of digits 1, 2, 3 with $n_1+2n_2+3n_3=11$ and $n_1+n_2+n_3=7$; the constraint reduces to $n_2+2n_3=4$, giving exactly 3 cases; count permutations for each.
$n_2+2n_3=4$ with $n_3\in\{0,1,2\}$:
**$(n_3=0,\,n_2=4,\,n_1=3)$:** $\dfrac{7!}{3!\,4!}=35$.
**$(n_3=1,\,n_2=2,\,n_1=4)$:** $\dfrac{7!}{4!\,2!\,1!}=105$.
**$(n_3=2,\,n_2=0,\,n_1=5)$:** $\dfrac{7!}{5!\,2!}=21$.
$|P|=35+105+21=161$.
Correct Answer: 4