# Math Help - Palindromes

1. ## Palindromes

I'm trying to find out the number of 7 digit numbers that are palindromes. My method is this (don't laugh):

$\frac{9}{10}*(\frac{10!}{6!}+(\frac{4!}{3!1!}+\fra c{4!}{2!2!}+\frac{4!}{2!1!1!}+\frac{4!}{4!})*\frac {10!}{6!4!})=8883$

If the correct number is 9000, what am I doing wrong? All wrong?

2. Originally Posted by billym
I'm trying to find out the number of 7 digit numbers that are palindromes. My method is this (don't laugh):

$\frac{9}{10}*(\frac{10!}{6!}+(\frac{4!}{3!1!}+\fra c{4!}{2!2!}+\frac{4!}{2!1!1!}+\frac{4!}{4!})*\frac {10!}{6!4!})=8883$
If the correct number is 9000, what am I doing wrong? ?
How many four digits (no leading zeros) numbers are there?
If you give one of those, say 6758, I can give you back a 7-digit palindrome: 6758576.

3. I'm not sure what you mean by that.

Here goes:

AAAA

1 * 9 = 9

AAAB
AABA
ABAA
BAAA
AABB
ABBA
ABAB

9 * 9 * 7 = 567

AABC
AACB
ABAC
ABCA
ACAB
ACBA

9 * 9 * 8 * 6 = 3888

ABCD

9 * 9 * 8 * 7 = 4536

9 + 567 + 3888 + 4536 = 9000

4. Originally Posted by billym
I'm not sure what you mean by that.
What an odd thing to say? Why not just ask what it means?

There is a one-to-one correspondence between 4-digit numbers and 7-digit palindromes.
There is a one-to-one correspondence between 5-digit numbers and 10-digit palindromes.

If n is even, there is a one-to-one correspondence between n-digit numbers and (2n-1)-digit palindromes.

If n is odd, there is a one-to-one correspondence between n-digit numbers and (2n)-digit palindromes.

5. Wasn't trying to sound like a jerk, I just wasn't sure what you were getting at at first.

I get the idea, though, that you just have to consider the first 4 digits and the first can't be a 0.

The first time round I was trying to do that: Find the number of selections of 4 without repetition that don't start with 0, and then add the number of selections with 4 repetitions, 3 repetitions, and 2 repetitons. My method was a bit ridiculous though (new to this).