SOLUTION: Determine the number of three figures numbers between 100 and 999 inclusive, have only two consecutive figures identical. (I don't really understand this question.)

Question 1182202: Determine the number of three figures numbers between 100 and 999 inclusive, have only two consecutive figures identical. (I don't really understand this question.)
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Perhaps part of your problem in understanding the problem is that the vocabulary and grammar are both poor....

Determine the number of numbers between 100 and 999 inclusive have only two consecutive identical.

Note also that "between 100 and 999 inclusive" is redundant, since it repeats the requirement that we are looking at only 3-digit numbers.

There are two kinds of 3-digit numbers that have only two consecutive digits the same: AAB and ABB.

(1) AAB (first two digits the same)

The first digit can be any of 9 digits (it can't be 0): 9 choices
The second digit has to be the same as the first: 1 choice
The third digit has to be different from the first two; and it can be 0: 9 choices

Number of 2-digit numbers of the form AAB: 9*1*9 = 81 (multiply the numbers of choices for each digit)

(2) ABB (last two digits the same)

The first digit can be any of 9 digits (it can't be 0): 9 choices
The second digit has to be different from the first; and it can be 0: 9 choices
The third digit has to be the same as the second: 1 choice

Number of 2-digit numbers of the form AAB: 9*9*1 = 81

ANSWER: The number of 3-digit numbers with only two consecutive digits the same is 81+81=162

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Thanks to tutor @ikleyn for catching the typo in my answer.

Answer by ikleyn(52784)   (Show Source): You can put this solution on YOUR website!
.

Small correction to the post by @greenestamps:

            ANSWER: The number of 3-digit numbers with only two consecutive digits the same is   81 + 81 = 162.

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