How many number of three digit numbers lying between 100

Leland Morrow

Leland Morrow

Answered question

2022-06-18

How many number of three digit numbers lying between 100 and 999 and having only two consecutive digits identical?

Answer & Explanation

trajeronls

trajeronls

Beginner2022-06-19Added 21 answers

Ask yourself first how many contain 00. Next, how many contain 11 (it's 17, you work it out). Then consider the number that contain 22, etc. You will get 9+17*9=162. I hope this helps.
boloman0z

boloman0z

Beginner2022-06-20Added 10 answers

In your comment, you said that the given solution was 9 • 9 + 1 • 9 + 8 • 9 = 162. I'll attempt to explain a logic that yields that calculation.
Consider the 3-digit numbers that start with two identical digits. There are 9 choices of the first digit (and inherently the second digit): 11, 22, 33, 44, 55, 66, 77, 88, and 99 (not 00 because the number is in the range 100-999). For each of these, there are 9 choices of the final digit (0-9, except whatever digit was already chosen for the first two). So, there are 9 • 9 such numbers.
Now, suppose that the number ends with two identical digits. There are 10 choices for the last digit (and inherently the second-to-last digit): 00, 11, 22, 33, 44, 55, 66, 77, 88, and 99, but we need to treat 00 separately from the rest. If the number ends with 00, then the first digit can be 1-9, so 9 choices, so 1 • 9. If the number ends with 11-99, there are 8 choices of first digit (1-9 except the digit already chosen), so 9 • 8.
While I have the 8 and 9 transposed in the final term, this is term-by-term the same expression as in the solution you gave.

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