Number of 9 digits numbers divisible by nine using the digits from 0 to 9 if each digit once is `K.8!` then K has the value equal to-

Explanation

As the number is divisible by `9`,

So, we can use the `9` digits from `0` to `8` or `1` to `9`.

When we are using digits `0` to `8`, the, `0` can not be the first number.

So,total numbers that can be formed `= 8*8*7*6*5*4*3*2*1 = 8*8!`

When we are using digits `1` to `9`, then,

Total numbers that can be formed `= 9!`

So, total numbers possible with both these two cases ` = 9!+8*8!`

`=9**8!+8*8! = 17*8!`

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