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Question 155008: hey could you please help me out with this permutation problem:
P(n,5)=20P(n,3): hey could you please help me out with this permutation problem:
P(n,5)=20P(n,3)
Answer by Edwin McCravy(2086) About Me  (Show Source):
You can put this solution on YOUR website!
P(n,5)=20P(n,3)

P(n,r)=n(n-1)(n-2)···(until_there_are_r_factors)

So, P(n,5)=n(n-1)(n-2)(n-3)(n-4)

and P(n,3)=n(n-1)(n-2)

Substituting,

n(n-1)(n-2)(n-3)(n-4)=20*n(n-1)(n-2)

Get a 0 on the right:

n(n-1)(n-2)(n-3)(n-4)-20*n(n-1)(n-2)=0

factor out n(n-1)(n-2)

n(n-1)(n-2)*((n-3)(n-4)-20)=0

Simplify the last parentheses:

n(n-1)(n-2)((n^2-7n+12)-20)=0

n(n-1)(n-2)(n^2-7n+12-20)=0

n(n-1)(n-2)(n^2-7n-8)=0

Factor the last parentheses:

n(n-1)(n-2)(n-8)(n+1)=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 > n, P(n,r) = 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