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Math Help - Palindromes

  1. #1
    Member billym's Avatar
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    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?
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  2. #2
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    Quote Originally Posted by billym View Post
    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.
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  3. #3
    Member billym's Avatar
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    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
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  4. #4
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    Quote Originally Posted by billym View Post
    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.
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  5. #5
    Member billym's Avatar
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    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).
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