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Math Help - let n >= 4 be an integer, prove..

  1. #1
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    Unhappy let n >= 4 be an integer, prove..

    let n >= 5 be an integer, prove..


    a) using algebra
    b) using combinatorial argument

    for part a) all you have to do is substitute 5 for n and you have:


    but for part b) i have to make up a story.. need help please...
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  2. #2
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    Quote Originally Posted by memee4eva View Post
    let n >= 5 be an integer, prove..


    a) using algebra
    b) using combinatorial argument

    for part a) all you have to do is substitute 5 for n and you have:


    but for part b) i have to make up a story.. need help please...
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  3. #3
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    Hello, memee4eva!

    Let n \geq 5 be an integer,

    prove: . {n\choose1} + 6{n\choose2} + 6{n\choose3} \:=\:n^3

    a) using algebra
    b) using combinatorial argument
    (a) You proved it true for n = 5 only.

    We have: . {n\choose1} + 6{n\choose2} + 6{n\choose3}

    That is: . n + 6\!\cdot\!\frac{n(n-1)}{2} + 6\!\cdot\!\frac{n(n-1)(n-2)}{6}

    . . . . = \;\;n + 3n(n-1) + n(n-1)(n-2)

    . . . . = \;\;n + 3n^2 - 3n + n^3 - 3n^2 + 2n

    . . . . =\;\; n^3


    And that is an algebraic proof.

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  4. #4
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    Hello memee4eva
    Quote Originally Posted by memee4eva View Post
    let n >= 5 be an integer, prove..


    a) using algebra
    b) using combinatorial argument

    for part a) all you have to do is substitute 5 for n and you have:


    but for part b) i have to make up a story.. need help please...
    For argument (b) here's the scenario: A password consists of 3 characters. How many possible passwords are there, if n characters are available to choose from, repetition being allowed?

    Method 1

    Each of the 3 positions in the password can be filled in n ways. There are therefore n^3 possible passwords.


    Method 2

    A If repetition is not allowed, there are \binom{n}{3} ways of choosing which 3 different characters to use; and 3! = 6 ways of arranging them to form the password. There are therefore 6\binom{n}{3} passwords where all three characters are different.

    B If two characters are repeated, and the third one is different, there are \binom{n}{2} ways of choosing which two characters will make up the password; then there are two ways of choosing which of these character is the one to be repeated; finally, there are then three ways of choosing the position to be occupied by the non-repeated character. So there are 6\binom{n}{2} passwords that use two different characters.

    C There are \binom{n}{1} ways of choosing a single character, and therefore \binom{n}{1} passwords made up of just one character.

    The total number of passwords is therefore \binom{n}{1} + 6\binom{n}{2} + 6\binom{n}{3}

    Grandad
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