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Math Help - Help with Solving equations involving factorial notation

  1. #1
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    Help with Solving equations involving factorial notation

    Hi, i need help solving for n

    a) n!/(n-2)!=930

    b) P(n,5)= 42 x P(n,3)

    Thank you
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  2. #2
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    n!/(n-2)!=n(n-1)=930
    Second degree equation
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  3. #3
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    would I then go on to solve?

    n(n-1)=930
    n2-n-930=0
    (n-30)(n+31)=0

    n=-30 or n=31
    n=31 as it cannot be negative

    thanks again
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by Carrick View Post
    Hi, i need help solving for n

    a) n!/(n-2)!=930

    b) P(n,5)= 42 x P(n,3)

    Thank you

    Use the Identity that

    n!=n(n-1)!=n(n-1)(n-2)!

    This yeilds

    \frac{n(n-1)(n-2)!}{(n-2)!}=930

    Simplify from here and solve for n.

    Hint: you will get quadratic equation and only keep the positive solution

    Try something similar for the nextone. Use the Formula for P(n,r)
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  5. #5
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    b) P(n,5)= 42 x P(n,3)

    Could someone let me know if i do this right, if not give me a little help?

    n!/n-5=42 x n!/n-3

    (n-1)(n-2)(n-3)(n-4)=42 x (n-1)(n-2)

    I'm stuck there, any help would be appreciated
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  6. #6
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    You can simplify by (n-1)(n-2) considering that 1 and 2 are not solutions of initial equation

    (n-3)(n-4)=42
    You can solve by using usual method of second degree equation or just looking at the equation
    n-3 and n-4 are two consecutive integers
    You surely know two consecutive integers whose product gives 42
    One is equal to n-3 and the other one to n-4
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