The number of seven-digit integers,with the s | vedclass

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Question

(C) Let the seven digits be $x_1, x_2, \dots, x_7$ where $x_i \in \{1, 2, 3\}$.We need to find the number of solutions to $x_1 + x_2 + \dots + x_7 = 1

Vedclass · Accepted Answer

$77$

Vedclass · Answer

(C) Let the seven digits be $x_1, x_2, \dots, x_7$ where $x_i \in \{1, 2, 3\}$.We need to find the number of solutions to $x_1 + x_2 + \dots + x_7 = 10$.This is equivalent to finding the coefficient of $x^{10}$ in $(x + x^2 + x^3)^7$.$(x + x^2 + x^3)^7 = x^7(1 + x + x^2)^7 = x^7 \left(\frac{1 - x^3}{1 - x}ight)^7 = x^7(1 - x^3)^7(1 - x)^{-7}$.We need the coefficient of $x^{10}$ in $x^7(1 - x^3)^7(1 - x)^{-7}$,which is the coefficient of $x^3$ in $(1 - x^3)^7(1 - x)^{-7}$.$(1 - x^3)^7(1 - x)^{-7} = (1 - 7x^3 + \dots)(1 + 7x + \frac{7 	imes 8}{2}x^2 + \frac{7 	imes 8 	imes 9}{6}x^3 + \dots)$.The coefficient of $x^3$ is $1 	imes \binom{7+3-1}{3} - 7 	imes 1 = \binom{9}{3} - 7 = 84 - 7 = 77$.Alternatively,the possible sets of digits are:$1, 1, 1, 1, 1, 2, 3$ (sum $= 10$): Number of arrangements $= \frac{7!}{5!} = 42$.$1, 1, 1, 1, 2, 2, 2$ (sum $= 10$): Number of arrangements $= \frac{7!}{4!3!} = 35$.Total $= 42 + 35 = 77$.

The number of seven-digit integers,with the sum of the digits equal to $10$ and formed by using the digits $1, 2,$ and $3$ only,is

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