How many 4-digit numbers are there where any two consecutive digits are different?

I know that there are $\binom{9}{1}$ ways to choose the first digit(from the left), and I think there are $\binom{10}{1}$ ways to choose the second, and third digit, and $\binom{9}{1}$ ways to choose the fourth digit since 2 digits must consecutive are different. I don't think I am going about this the right way though because if you multiply $\binom{10}{1}*\binom{10}{1}*\binom{9}{1}*\binom{9}{1}$ which seems way too big. Can someone help me go about this?

  • 2
    $\begingroup$ 9 choices for each digit...think about why $\endgroup$ – Don Thousand Mar 22 at 3:50
  • $\begingroup$ It's more usual to write $\binom 91$ as $9$, etc. $\endgroup$ – Lord Shark the Unknown Mar 22 at 4:02

There are 9 digits to choose from for the first digit (most significant digit). Why do you reckon there are 10 digits to fill the second spot? There are still only 9 digits that can fill it (0-9 except the digit that is in the first spot). This is also true for every digit after that too.

In fact, by my reckoning there must be ${9 \choose 1}^4$ ways form 4 digit numbers with not two consecutive digits.


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