SOLUTION: hey could you please help me out with this permutation problem: P(n,5)=20P(n,3)

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Question 155008: hey could you please help me out with this permutation problem:
P(n,5)=20P(n,3)
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
P%28n%2C5%29=20P%28n%2C3%29

P%28n%2Cr%29=n%28n-1%29%28n-2%29···%28until_there_are_r_factors%29

So, P%28n%2C5%29=n%28n-1%29%28n-2%29%28n-3%29%28n-4%29

and P%28n%2C3%29=n%28n-1%29%28n-2%29

Substituting,

n%28n-1%29%28n-2%29%28n-3%29%28n-4%29=20%2An%28n-1%29%28n-2%29

Get a 0 on the right:

n%28n-1%29%28n-2%29%28n-3%29%28n-4%29-20%2An%28n-1%29%28n-2%29=0

factor out n%28n-1%29%28n-2%29

n%28n-1%29%28n-2%29%2A%28%28n-3%29%28n-4%29-20%29=0

Simplify the last parentheses:

n%28n-1%29%28n-2%29%28%28n%5E2-7n%2B12%29-20%29=0

n%28n-1%29%28n-2%29%28n%5E2-7n%2B12-20%29=0

n%28n-1%29%28n-2%29%28n%5E2-7n-8%29=0

Factor the last parentheses:

n%28n-1%29%28n-2%29%28n-8%29%28n%2B1%29=0

Setting each factor = 0, we get the
solutions:

n=0, n=1, n=2, n=8, n=-1.

The first three are considered correct,
because when r+%3E+n, P%28n%2Cr%29+=+0

And n=8 is a solution.  But the last one
n=-1 is not a solution because permutations
are not defined for negative numbers.

So the four solutions are

n=0, n=1, n=2, n=8

Possibly your teacher may want only
the solution n=8; however, the first 
three are solutions since they cause 
both sides of the original equation
to equal 0.

Edwin