Let P be the set of seven-digit numbers with | vedclass

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(B) Let the digits be $x_1, x_2, x_3, x_4, x_5, x_6, x_7 \in \{1, 2, 3\}$ such that $\sum_{i=1}^{7} x_i = 11$. Let $n_1, n_2, n_3$ be the number of ti

Vedclass · Accepted Answer

$161$

Vedclass · Answer

(B) Let the digits be $x_1, x_2, x_3, x_4, x_5, x_6, x_7 \in \{1, 2, 3\}$ such that $\sum_{i=1}^{7} x_i = 11$. Let $n_1, n_2, n_3$ be the number of times $1, 2,$ and $3$ appear respectively. Then $n_1 + n_2 + n_3 = 7$ and $1n_1 + 2n_2 + 3n_3 = 11$. Subtracting the first from the second: $n_2 + 2n_3 = 4$. Case $1$: If $n_3 = 0$,then $n_2 = 4$ and $n_1 = 3$. The number of permutations is $\frac{7!}{3!4!0!} = 35$. Case $2$: If $n_3 = 1$,then $n_2 = 2$ and $n_1 = 4$. The number of permutations is $\frac{7!}{4!2!1!} = 105$. Case $3$: If $n_3 = 2$,then $n_2 = 0$ and $n_1 = 5$. The number of permutations is $\frac{7!}{5!0!2!} = 21$. Total number of elements in $P = 35 + 105 + 21 = 161$.

Let $P$ be the set of seven-digit numbers with the 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 the set $P$ is:

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