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?


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

1 Answer 1


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.

Simple test: write a simple computer program that enumerates all of them. See who's right...

  • $\begingroup$ I noticed you went to York @HennoBrandsma. I'm currently at U of T myself. Thanks for the help $\endgroup$ Commented Dec 5, 2015 at 19:05
  • $\begingroup$ @dc3rd I went to university in the Netherlands (but I moderate a site at York, indeed. Been to Toronto only once. $\endgroup$ Commented Dec 5, 2015 at 19:08
  • $\begingroup$ interesting, how did that come about? $\endgroup$ Commented Dec 5, 2015 at 19:09
  • $\begingroup$ I was a big contributor in the beginning, so I was asked to help moderate. I visited Toronto once for holidays before a conference up in North Bay. $\endgroup$ Commented Dec 6, 2015 at 17:32

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