Permutation Question - Solver for "n"

**cnmath16** · November 23rd 2008, 10:34 AM

Solve for n where....

b) P (n,5) = 42 x P(,3) -- out of 5 marks

I know that you have to split up the equation into two sides (left side = right side)

This is as far as I've come with the question..

LS = n! divided by (n-5) RS = 42 x n! divided by 2!

please help me with the rest of the steps!

Thanks in advance

**Plato** · November 23rd 2008, 11:55 AM

$\frac{{n!}}<br /> {{\left( {n - 5} \right)!}} = 42\frac{{n!}}<br /> {{\left( {n - 3} \right)!}} \Rightarrow \quad \frac{{\left( {n - 3} \right)!}}<br /> {{\left( {n - 5} \right)!}} = 42$
Can you solve it now?

**cnmath16** · November 23rd 2008, 12:14 PM

From the book I am learning from, it shows separating into two equations: left side of the equal sign and right side of the equal sign. Are you combining them together?

I know how to break down (n-3)! into n(n-1)(n-2).. but I'm not sure how to break down (n-5)!

**o_O** · November 23rd 2008, 12:24 PM

We're just working with a normal, algebraic equation:
.... $\begin{aligned} P(n,5) & = 42 \cdot P(n,3) \\ \frac{n!}{(n-5)!} & = 42 \cdot \frac{n!}{(n-3)!} \\ \frac{(n-3)!}{(n-5)!} & = 42\end{aligned}$

I know how to break down (n-3)! into n(n-1)(n-2).. but I'm not sure how to break down (n-5)!

No ... $(n-3)! = (n-3)(n-4)(n-5)(n-6) \cdots (2)(1)$ .

Remember, you're multiplying all the numbers from (n-3) all the way down to 1.

For example, if $n = 15$ , we would have: $\underbrace{(n-3)!}_{12!} = \underbrace{(n-3)}_{12}\underbrace{(n-4)}_{11}\underbrace{(n-5)}_{10}\underbrace{(n-6)}_{9} \cdots (2)(1)$

For your question, expand $(n-5)!$ in the denominator as you would with $(n-3)!$ in the numerator and see what cancels

**cnmath16** · November 23rd 2008, 12:57 PM

Thank you..
So am I correct in saying the break down for (n-3)! is n(n-2) (n-1)

**o_O** · November 23rd 2008, 01:00 PM

No ... Take for example n = 6. $(6-3)! = 3! = 6$ but $(6)(6-2)(6-1) = 120$

Read the above post again carefully.

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