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Math Help - Simplify the ratio

  1. #1
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    Simplify the ratio

    simplify the ratio
    (n+4)! / (n+3)!


    ok i know how to simplify ratios like 4!/3! which = 4 but when they put the n in the problem i dont know what to do.
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  2. #2
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    Quote Originally Posted by flexus View Post
    simplify the ratio
    (n+4)! / (n+3)!


    ok i know how to simplify ratios like 4!/3! which = 4 but when they put the n in the problem i dont know what to do.
    Just like 5!=(5)(4)(3)(2)(1), we can use the same rationale here;

    (n+4)!=(n+4)(n+4-1)(n+4-2)...

    \frac{(n+4)!}{(n+3)!} = \frac{(n+4)(n+3)(n+2)(n+1)...}{(n+3)(n+2)(n+1)...}

    The multiplications cancel as the terms lower to the right, can you see what might be left over?
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  3. #3
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    Remember what a factorial is and write it out, then cancel.

    Like so:

    (n+4)!=(n+4)(n+3)(n+2)(n+1)(n)............

    (n+3)!=(n+3)(n+2)(n+1)(n).................

    So, \frac{(n+4)!}{(n+3)!}=\frac{(n+4)(n+3)(n+2)(n+1)(n  )............}{(n+3)(n+2)(n+1)(n).................  }

    See the cancellations?. You are left with n+4.
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  4. #4
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    Quote Originally Posted by Kasper View Post
    Just like 5!=(5)(4)(3)(2)(1), we can use the same rationale here;

    (n+4)!=(n+4)(n+4-1)(n+4-2)...

    \frac{(n+4)!}{(n+3)!} = \frac{(n+4)(n+3)(n+2)(n+1)...}{(n+3)(n+2)(n+1)...}

    The multiplications cancel as the terms lower to the right, can you see what might be left over?
    thank you i see what you did. n+4
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