Question

Total number of $6$ digit numbers that can be formed, having the property that every succeeding digit is greater than the preceding digit, is equal to

• A. ${}^9{C}_3$
• B. ${}^{10}{C}_3$
• C. ${}^9{P}_3$
• D. ${}^{10}{P}_3$

Answer 1

The correct option is A ${}^9{C}_3$

Let the six digit number be $x_1{x}_2{x}_3{x}_4{x}_5{x}_6$ such that $x_1<x_2<x_3<x_4<x_5<x_6$

Clearly no digit can be zero. Also, all the digits are distinct. So, let us first select six digits from the list of digits $1,2,3,4,5,6,7,8,9$ which can be done in ${}^9{C}_6$ ways.

After selecting these digits they can be put only in one order.

Thus, total number of such numbers is ${}^9{C}_6\times 1={}^9{C}_3$.