# 7 digit number combinations where the first 3 digits must be equal to another 3.

Seven-digit telephone numbers are not allowed to begin with $$0$$ or $$1$$.

I can only remember a seven-digit telephone number if the first three digits (the "prefix") are equal to either the next three digits or the last three digits. For example, I can remember $$389$$-$$3892$$ and $$274$$-$$9274$$.

How many seven-digit telephone numbers can I remember?

I split this up into 2 cases, 1 where the first 3 is the same as the next 3 and another where the first three is the same as the last 3 and I got 8000 for each case and multiplied by 2 which is 16000. However, there are cases where all the digits are the same, like 8888888, but I don't know how many there are and what to subtract from 16000.

• The only cases in which both occur are those when every digit is the same as you have seen. There are only $8$ of these cases so the answer should be $16000-8=15992$. Commented Jun 22, 2020 at 15:44

First I would check how many "first three digits" there are, which are not allowed to begin with $$0$$ or $$1$$. So there are $$8 \times 10 \times 10 = 800$$ possible "first three digits"
Pick one possibility, call it $$ABC$$, and split it into cases as you said.
Case "ABC-ABCx": well there are 10 choices for the last digit, totalling $$800 \times 10 = 8000$$ possibilities
Similarly for case "ABC-xABC". Hence so far we have $$Case1 + Case2 = 16,000$$
But as you point out we overcounted, since "ABC-ABCx"="ABC-xABC" exactly when $$x=A,x=C$$ and $$B=A$$. Namely phone numbers of the form "AAA-AAAA", of which there are $$8$$ possibilities. So final answer is $$Total = 16,000 - 8$$