Question 1182202
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Perhaps part of your problem in understanding the problem is that the vocabulary and grammar are both poor....<br>
Determine the number of {{{cross(three)}}} {{{cross(figures)}}} {{{red(three-digit)}}} numbers between 100 and 999 inclusive {{{red(that)}}} have only two consecutive {{{cross(figures)}}} {{{red(digits)}}} identical.<br>
Note also that "between 100 and 999 inclusive" is redundant, since it repeats the requirement that we are looking at only 3-digit numbers.<br>
There are two kinds of 3-digit numbers that have only two consecutive digits the same: AAB and ABB.<br>
(1) AAB (first two digits the same)<br>
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<br>
Number of 2-digit numbers of the form AAB: 9*1*9 = 81  (multiply the numbers of choices for each digit)<br>
(2) ABB (last two digits the same)<br>
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<br>
Number of 2-digit numbers of the form AAB: 9*9*1 = 81<br>
ANSWER: The number of 3-digit numbers with only two consecutive digits the same is {{{cross(81+81+162)}}} 81+81=162<br>
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Thanks to tutor @ikleyn for catching the typo in my answer.<br>