Q: Find out how many 3 figure numbers, lying between 100 and 999 inclusive, have 2 and only 2 consecutive figures identical.
Thank you for helping! ^^
You can choose $\displaystyle \frac{10!}{2! \cdot (10-2)!} $ sets of two numbers and you can permute them 2 times.
If you pick eg 2 and 9, you can make 992 and 229.
You need to consider that 0 and 1 can only make 100 so do -10 at the end.
I think you should get 80 numbers in the end.
I just realize I'm wrong. Sorry!
Let's do it like this.
You have 9 possibilities for first number and 9 for the second when the second is repeated. 9 x 9 = 81
You have 9 possibilities for first number and 9 for the second when the first is repeated. 9 x 9 = 81
Way more simple and works well. Less esthetic although.