# Suppose campus telephone numbers consist of any four digits

Suppose campus telephone numbers consist of any four digits, how many ways can a campus telephone number contain exactly two different digits?

Soln: I broke it into two cases: (3,1) and (2,2)

(3,1): $$\binom{10}{2} \binom{4}{3} X 2$$

(2,2): $$\binom{10}{2} \binom{4}{2}$$

But the solution is:

(3,1): $$\binom{10}{2} \binom{4}{3}$$

(2,2): $$\binom{10}{2} X 3$$

In (3,1) I though that either of the two selections could be the digit with 3 occurances of it in the set so I multiplied by 2 to account for that. In the (2,2) case I thought that if 2 of the 4 places for digits were selected then by default the other places are accointed for. What was flawed in my thought process?

Thanks

• Your reasoning is flawless. Challenge the author of the other solution. Commented Dec 5, 2015 at 18:31

In the $(3,1)$ case we pick position of the odd one out number, so we have $4 = \binom{4}{1}$ choices for that, pick the digit we want there, so times 10, and then from the remaining 9 digits also pick one to put on the remaining places, so $4 \times 10 \times 9$. This agrees with your $(3,1)$ answer as well.
For $(2,2)$: pick two digits and two places, so $\binom{10}{2}\binom{4}{2}$. By symmetry we are done, because we can interchange places and digits to get the same result. So I agree with your reasoning here as well.