1. Permutations

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

Solve each equation for "n"
$\displaystyle \frac{(n+1)!}{(n-2)!} =20(n-1)$

$\displaystyle _nP_3=24$

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

2. Originally Posted by nightrider456
been stuck on these for a while now, any help would be great.

Solve each equation for "n"
$\displaystyle \frac{(n+1)!}{(n-2)!} =20(n-1)$

$\displaystyle _nP_3=24$

$\displaystyle \frac{n!}{84}=_{n-2}P_{n-4}$
1) Note that (n+1)! = (n+1)n(n-1)(n-2)! Substitute, simplify and solve the equation.

For the other two questions, start by substituting the definitions and simplifying. What do you get? Where are you still stuck?

3. 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.