If all the six-digit numbers x_1 x_2 x_3 x_4 | vedclass

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(C) We are looking for six-digit numbers $x_1 x_2 x_3 x_4 x_5 x_6$ such that $1 \le x_1 < x_2 < x_3 < x_4 < x_5 < x_6 \le 9$. The number of such integ

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) We are looking for six-digit numbers $x_1 x_2 x_3 x_4 x_5 x_6$ such that $1 \le x_1 To find the $72^{	ext{th}}$ number,we count how many numbers start with specific digits:- Numbers starting with $1$: $\binom{8}{5} = 56$ numbers.- Numbers starting with $23$: $\binom{6}{4} = 15$ numbers.Total numbers counted so far: $56 + 15 = 71$.The $71^{	ext{st}}$ number is the last number starting with $23$,which is $235678$.The $72^{	ext{nd}}$ number is the first number starting with $24$,which is $245678$.The sum of the digits is $2 + 4 + 5 + 6 + 7 + 8 = 32$.

If all the six-digit numbers $x_1 x_2 x_3 x_4 x_5 x_6$ with $0 < x_1 < x_2 < x_3 < x_4 < x_5 < x_6$ are arranged in increasing order,then the sum of the digits in the $72^{\text{th}}$ number is $............$.

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