Question

Total number of six-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

Answer: (A)

${}^9{C}_3$

$x_1<x_2<x_3<x_4<x_5<x_6$, when the number is $x_1{x}_2{x}_3{x}_4{x}_5{x}_6$.
Clearly
No digit can be zero. Also, all the digits are distinct. So selection of six digits from the list of digits $1,2,3,4,5,6,7,9$ can be done in ${}^9{C}_6{=}^9{C}_3$ ways.