Question

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

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

Answer 1

Answer: (C)

We have to form six digit number $x_1{x}_2{x}_3{x}_4{x}_5{x}_6$ having $x_1<x_2\le x_3<x_4<x_5<x_6$
So here arises following cases:
Case I : $x_1<x_2<x_3<x_4<x_5<x_6$
Now single order is already fixed so we just neet to select 6 numbers out of 9 possible numbers $\{1,2,3,4$, $5,6,7,8,9\}$
zero is neglected because it can't be placed at any place
This can be done in ${}^9{C}_6$ ways
Case II : $x_1<x_2=x_3<x_4<x_5<x_6$
Now we have to select 5 digit
No. of ways $={}^9{C}_5$
Case III : $x_1<x_2=x_3<x_4<x_5=x_6$
Now we have to select 4 digits done in ${}^9{C}_4$ ways.
Case IV: $x_1<x_2<x_3<x_4<x_5=x_6$
We have select 5 digits
which is done in $={}^9{C}_5$ ways.
Total no. ways $={}^9{C}_6+2\cdot {}^9{C}_5+{}^9{C}_4={}^9{C}_6+{}^9{C}_5+{}^9{C}_5+$ ${}^9{C}_4={}^{10}{C}_6+{}^{10}{C}_5={}^{11}{C}_6$