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If all the six digit numbers \mathrm{x_1 x_2 x_3 x_4 x_5 x_6}  with \mathrm{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^{\text {th }}} number is
 

Option: 1

32


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Number of six digit number starting with 1 is 1 \ldots \ldots \ldots \ldots . .{ }^8 \mathrm{C}_5=56
As remaining five digits can be selected from 8 digits that are greater than (i.e., 2, 3, 4, 5, 6, 7, 8)
Number of six digit number starting with 23 \ldots \ldots \ldots={ }^6 \mathrm{C}_4=15
                                                        \mathrm{Total =56+15=71}
Now, \mathrm{72^{\text {nd }} number =245678}
\therefore sum of the digits =2+4+5+6+7+8=32

Posted by

Deependra Verma

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