Class 11

Math

Algebra

Permutations and Combinations

The number of $n$ digit numbers such that no two consecutive digits are same is

- $9!$
- $n_{9}$
- $9_{n}$
- $9n$

**Correct Answer: ** Option(c)

**Solution: **There are ${0,1,2,.....10},$ $10$ different digits of which $1_{st}$ place is filled in $9$ ways, $2_{nd}$ place in $9$ ways (since $0$ can be included and first digit is not repeated) and so on $n$ times.Thus number of ways is $=9×9×............=9_{n}$