As the series lies between 100 & 1000 , the desired series of palindrome will have only 3 digit nos.
Start from the 1 series, keeping the 1st & last digit as 1 (1st no. will be 101) ,
we can have 10 (0,1,2...9) different ways for choosing the 2nd digit .
Repeating the same process for 2 series (where 1st & last digit is 2) and so on...we will always have 10 different nos. for each series.
Therefore , adding the nos. till the 9 series (last no. being 999) = 10*9 = 90

Note, all integers between 100 and 1000 will have 3 digits.
Take the task of creating palindromes and break it into stages:

Stage 1: Select a digit to be the hundreds digit AND the units digit
NOTE: In order for the resulting integer to be a palindrome, the units digit must be the SAME as the hundreds digit
The hundreds digit (and units units digit) can be 1,2,3,4,5,6,7,8, or 9
So, this stage can be completed in 9 ways.

Stage 2: Select the tens digit.
This digit can be 0,1,2,3,4,5,6,7,8, or 9
So, this stage can be completed in 10 ways.

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a 3-digit palindrome) in (9)(10) ways
Answer:

Cheers,
Brent
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Stage 1 - Selecting from 9 digits I get (0-9)
Stage 2 - Selecting from 10 digits - Little lost here . Since one digit is already selected in stage 1 - Shouldnt it be 9 digits to select from ?

Digits such as 999 and 444 are also palindromes.

So for the first place there are 9 options namely (1,2,3,4,5,6,7,8,9) and second digit has 10 options (0-9).
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There are two situations:

1. xyx (ex: 101, 121..)
2. xxx (ex: 111, 2222..)

For the first case:

xyx: The hundredth position can be taken by any one among the nine digits..(1,2,3...9)
The tenth position can be any one among the nine numbers (0,1,2..9 except the digit which we have taken in the hundredth position)
The unit's place should be the same digit which we assigned in hundredth position.
Hence this can be arranged in 9 * 9 * 1 = 81

For the second case:

xxx: The hundredth position can be taken by any one among the nine digits..(1,2,3...9)
The tenth and the units's place should take the same digit which we assigned in hundredth position.
Hence this can be arranged in 9 * 1 * 1 = 9

The total number of palindromes between 100 and 1000 = 81 + 9 =90.

ANSWER: 90

From 100 - 200 there is 10 palindromes. Those are 101,111,121,131,141,151,161,171,181,191.
As we can see when the first and third digit matches, it is a palindrome.
Similarly 201-300 there is 10 Palindromes.
301-400 there is 10 palindromes.
So in total 100-1000 there is 10*9 = 90 palindromes.

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