[Math] Number of 4-digit numerals with at most 2 different digits

combinatorics

How many 4 digit numbers are there which contain not more than two different digits?

As usual, I will highlight my attempt:

  1. I can have 9 numbers using only 1 digit, e.g. $1111, 2222, \dots 9999$ = 9 digits

  2. Using two digits out of 9 (excluding 0) (selected in $\,^{8}C_{2}$ ways), I can arrange them in $4!$ ways. Furthermore, I can use one digit out of the selected 2 once or thrice (e.g. $2111$ or $2221$) or simply both digits 2 times (e.g. $2288$, $6677$). Hence I get: $$\,^{8}C_{2} \times \left(\frac{4!}{2!\times2!} + \frac{4!}{3! \times 1!}\times 2\right) = 392$$

  3. Finally, I consider numbers with zeroes. With zero selected as a digit, I have 9 options for other digits. And using the same logic as in 2) I can have either 1 zero ($8880$), 2 zeroes (e.g. $8800$) or 3 zeroes in a number ($7000$).
    This gives me $$9\times \left( \frac{4!}{2!\times2!} + \frac{4!}{3!\times1!} + 1\right) = 99$$

the above 3 steps yield a total number of $9+392+99=500$

However, the answer i have with me is $576$.

Best Answer

André has come up with a nice and simple alternative solution, but I'm going to try to salvage your method, which is essentially correct.

  1. This step does indeed give you 9 numbers.
  2. There are 9, not 8, numbers between 1 and 9, so you want to use $9 \choose 2$ not $8 \choose 2$. Indeed, $${9 \choose 2} \times \left(\frac{4!}{2!\times2!} + 2\times\frac{4!}{3!\times1!}\right) = 504$$
  3. You've counted too many here, because you've not excluded putting the zero first. There are actually only three ways of having one zero (8880, 8808, 8088) and three of having two (8800, 8080, 8008). Hence the total number of ways, including the one way of having three zeroes, is $9 \times (3 + 3 + 1) = 63$.

Now $9 + 504 + 63 = 576$. So you had the right approach, but you just messed up some of the arithmetic.

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