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 the increasing order, then the sum of the digits in the $\mathrm{72^{th}}$ number is _____________.
Answer (integer)
32
Solution
$1 \ldots \ldots \ldots \ldots \ldots \rightarrow{ }^{8} C_{5}=56$
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23 $\ldots\ldots\ldots\ldots\ldots\rightarrow{ }^{6} C_{4}=\frac{15}{71}$
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$72^{\text {th }}$ number $=245678$
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Sum $=32$
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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