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Math Help - Permutation Question - Solver for "n"

  1. #1
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    Permutation Question - Solver for "n"

    Solve for n where....

    b) P (n,5) = 42 x P(,3) -- out of 5 marks

    I know that you have to split up the equation into two sides (left side = right side)

    This is as far as I've come with the question..

    LS = n! divided by (n-5) RS = 42 x n! divided by 2!

    please help me with the rest of the steps!

    Thanks in advance
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  2. #2
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    \frac{{n!}}<br />
{{\left( {n - 5} \right)!}} = 42\frac{{n!}}<br />
{{\left( {n - 3} \right)!}} \Rightarrow \quad \frac{{\left( {n - 3} \right)!}}<br />
{{\left( {n - 5} \right)!}} = 42
    Can you solve it now?
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  3. #3
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    From the book I am learning from, it shows separating into two equations: left side of the equal sign and right side of the equal sign. Are you combining them together?

    I know how to break down (n-3)! into n(n-1)(n-2).. but I'm not sure how to break down (n-5)!
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  4. #4
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    We're just working with a normal, algebraic equation:
    .... \begin{aligned} P(n,5) & = 42 \cdot P(n,3) \\ \frac{n!}{(n-5)!} & = 42 \cdot \frac{n!}{(n-3)!} \\ \frac{(n-3)!}{(n-5)!} & = 42\end{aligned}

    I know how to break down (n-3)! into n(n-1)(n-2).. but I'm not sure how to break down (n-5)!
    No ... (n-3)! = (n-3)(n-4)(n-5)(n-6) \cdots (2)(1).

    Remember, you're multiplying all the numbers from (n-3) all the way down to 1.

    For example, if n = 15, we would have: \underbrace{(n-3)!}_{12!} = \underbrace{(n-3)}_{12}\underbrace{(n-4)}_{11}\underbrace{(n-5)}_{10}\underbrace{(n-6)}_{9} \cdots (2)(1)

    For your question, expand (n-5)! in the denominator as you would with (n-3)! in the numerator and see what cancels
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  5. #5
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    Thank you..
    So am I correct in saying the break down for (n-3)! is n(n-2) (n-1)
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    No ... Take for example n = 6. (6-3)! = 3! = 6 but (6)(6-2)(6-1) = 120

    Read the above post again carefully.
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