# Help with Solving equations involving factorial notation

• Nov 28th 2008, 08:50 AM
Carrick
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
• Nov 28th 2008, 08:56 AM
running-gag
n!/(n-2)!=n(n-1)=930
Second degree equation
• Nov 28th 2008, 09:08 AM
Carrick
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 AM
TheEmptySet
Quote:

Originally Posted by Carrick
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 $P(n,r)$
• Dec 1st 2008, 08:17 AM
Carrick
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 AM
running-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