Palindromes Are Your Best Pal

A palindrome of 5 digits contains 3 unique digits. to find the amount of combinations of three digits is 10^3. but because you can not start with 0 in the hundreds place you remove 100 unique combinations leading to

10^3 -100 = 900 there are 900 unique palindrome

*if anyone can explain this better please do.

The 5-digit palindrome will be in the form of $\overline{abcba}$ , each letter representing one (not necessarily distinct) digits

To make the 5-digit number, the leading digit shall not be zero. This make the first digit selection being 9 method. For $b$ and $c$ , since it is not in the leading digit, it can be picked freely from 0 to 9, for 10 method each digit

Thus, the possible combination being $9\times10\times10 = 900$ 5-digit palindromes

Given a 5 digits palindrome ABCBA, we can make a 3 digits number ABC, and vice versa (one to one corresponding).

Therefore the answer is the number of 3 digits numbers: 999-99=900.

([999 positive numbers of 3 digits or less] minus [99 of them which have 2 digits or less])

Consider 5 boxes no of ways of filling the first box=9(0 not ) second=10 third =10. And then the no repeat so no of ways of filling the next 2 boxes is 1 on multiplying 9 10 10=900

Let ABCBA be a 5 digit palindrome.

Because A cannot be $0$ , it can take the values $1-9$ . B and C can be $0$ so can be any value $0-9$ .

Hence the total number is $9\times10\times10=900$

The digits can be looked at like this: 12 3 45

4 and 5 don't matter. So it comes down to how many 2 digit numbers that exist, multiplied by how many 1 digit numbers that exist.

2 digits (10 through 99)

99-9=90

1 digit (0 through 9)

9+1 for the zero=10

So:

90×10=900

_ _ _ _ _ for first place 9 digits last filled second place 10 digits fourth filled and third 10 digits. Total=9.10.10=900.

We are only concerned with the first three digits in order for the number to be a palindrome. There are 9 possible unique digits for the first digit (1-9 since you can't start the number with a 0), 10 possible unique digits for the second digit (0-9), and 10 possible digits for the third digit (0-9).

Let's start the number with the first and second digits being 10 * , then by default the last two digits must mirror to become 10*01 for the number to be a palindrome. Then the third digit can be one of the 10 possible digits from 0-9. This means there are 10 possible combinations for this rule.

Now, let's cycle the second digit through one of the other 9 possible digits (1-9). So now there are 9x10=90, but we add 10 from above to yield 100 possible combinations for the number, if the first digit is 1.

But there are 9 possible unique digits for the first digit, so we multiply 100x9=900 combinations.

For a palindrome of 5 digits, the number of ways to choose the first and last digit is 9, since they are the same and cannot be 0. The number of ways to choose the thousands and tens digit is 10. The number of ways to choose the middle digit is 10. Hence, the answer is 9 10 10=900.