Hi, i need help solving forn

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

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

Thank you

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- Nov 28th 2008, 08:50 AMCarrickHelp 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 - Nov 28th 2008, 08:56 AMrunning-gag
n!/(n-2)!=n(n-1)=930

Second degree equation - Nov 28th 2008, 09:08 AMCarrick
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 - Nov 28th 2008, 10:12 AMTheEmptySet

Use the Identity that

$\displaystyle n!=n(n-1)!=n(n-1)(n-2)!$

This yeilds

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

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 $\displaystyle P(n,r)$ - Dec 1st 2008, 08:17 AMCarrick
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? (Happy)

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 - Dec 1st 2008, 09:09 AMrunning-gag
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