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

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?

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

## 1 Answer

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.