How many different numbers can be found using 6 out of 8 digits $12345678$ beginning with $21$?

Question

How many different numbers can be found using $6$ out of $8$ digits $1,2,3,4,5,6,7,8$, if the numbers are not repeated but must begin with the digits $21$?

I'm getting 360. Am I correct?

$2,1,3,4,5,6 $

$$ 6!/(6-4)!=360 $$

Answer 1

Yes, you are correct--it's the number of permutations of $4$ objects chosen from a set of $6$ ($3,4,5,6,7,8$).

Answer 2

I don't know how you arrived at your formula, but I get the same answer.

The first two digits are fixed, so we only have to count the number of ways to fill up the remaining $4$ digits. We have $6$ digits to choose from (the digits $1$ through $8$, but not $1$ and $2$ because they've already been used up). So there are $6\choose4$ possible combinations of digits, but that doesn't take into account the differnet ways they can be ordered, so we multiply by $4!$ and indeed get the answer $360$.