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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

If a five-digit number divisible by $3$ is to be formed using the digits $0, 1, 2, 3, 4,$ and $5$ without repetition,then the total number of ways this can be done is:

If $_n{P_4} = 24 \times \binom{n}{5}$,then $n = \dots$

The number of words that do not start and end with vowels,formed using all the letters of the word $'UNIVERSITY'$ such that all vowels are in alphabetical order,is

Let $S = \{0, 1, 2, 3, \ldots, 100\}$. The number of ways of selecting $x, y \in S$ such that $x \neq y$ and $x + y = 100$ is

Let $S$ be the set of all permutations $a_1, a_2, \ldots, a_6$ of $1, 2, \ldots, 6$ such that $a_1, a_2, \ldots, a_k$ is not a permutation of $1, 2, \ldots, k$ for any $k, 1 \leq k \leq 5$. Then the number of elements in $S$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module