been stuck on these for a while now, any help would be great.

Solve each equation for "n"

$\displaystyle

\frac{(n+1)!}{(n-2)!} =20(n-1)

$

$\displaystyle

_nP_3=24

$

$\displaystyle

\frac{n!}{84}=_{n-2}P_{n-4}

$

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- Mar 26th 2010, 10:35 PMnightrider456Permutations
been stuck on these for a while now, any help would be great.

Solve each equation for "n"

$\displaystyle

\frac{(n+1)!}{(n-2)!} =20(n-1)

$

$\displaystyle

_nP_3=24

$

$\displaystyle

\frac{n!}{84}=_{n-2}P_{n-4}

$ - Mar 26th 2010, 10:43 PMmr fantastic
- Mar 26th 2010, 10:53 PMANDS!
If the answers aren't popping out, sometimes it helps to test some values:

$\displaystyle

\frac{(n+1)!}{(n-2)!} =20(n-1)

$

Suppose n=4. Then:

$\displaystyle \frac{5*4*3*2*1}{2*1}=20(3)$

It looks like we get some common factors in the numerator and denominator on the LHS (RHS isnt important right now). In general though we have:

$\displaystyle \frac{(n+1)(n)(n-1)*(n-2)!}{(n-2)!}=20(n-1)$

Factoring:

$\displaystyle (n+1)(n)(n-1)=20(n-1) \Rightarrow (n+1)(n)=20$

You should be able to take it from there - hopefully. :O

2) Use the definition:

$\displaystyle

_nP_3=24 \Rightarrow \frac{n!}{(n-3)!}=24

$

We use the EXACT same technique as above:

$\displaystyle \frac{n!}{(n-3)!}=24 \Rightarrow \frac{(n)(n-1)(n-2)*(n-3)!}{(n-3)!}=24$

Now solve.

See if you can use the techniques we used in the first two to solve the last one. First write out the problem using the definition. Then see if things start simplifying.