Question

The total numeber of six digit numbers $x_1{x}_2{x}_3{x}_4{x}_5{x}_6$ having the property that $x_1<x_2\le x_3<x_4<x_5\le x_6$ is equal to

• A. ${}^{10}{C}_6$
• B. ${}^{12}{C}_6$
• C. ${}^{11}{C}_6$
• D. none of these

Answer 1

The correct option is A ${}^{10}{C}_6$

We know that the first digit must be 1 or higher and last digit must be 9 or lower

Let's also imagine we have 8 balls each labeled with $a+1$ sign

Finally imagine each number as a divider

$1\le x_1<x_2{\le x}_3{<x}_4{<x}_5{\le x}_6\le 9......|.....|......|.....|.....|.....|.....$

we want to distribute those $8$ balls into the spaces betwee3n the dividers.That will tell us how much to add to the prior digit to get to the next digit.

But first we have to put $a+1$ ball under each of the $<$ signs,because difference between those numbers has to be at least 1

$1\le x_1<x_2{\le x}_3{<x}_4{<x}_5{\le x}_6\le 9.....|0|....|0|0|.....|.....$

That leaves $5$ balls we need to place between 6 dividers

so we have 5 balls and 6 dividers that's total $11$ objects

$11$ objects to be picked up for 6 dividers

so we get ${{}^{11}C}_6$