We are to solve for n

Please verify

P(n,2)=15

My solution for this problem

By factoting

n(n-1) = 15

n=15 ; n= 16

Another Problem

P(n,5) = 20 P(n,3)

I have no idea how to do it

Hoping for replies, thanks

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- Jan 13th 2009, 02:43 AMImKoA Permutation Problem-Urgent
We are to solve for n

Please verify

**P(n,2)=15**

My solution for this problem

By factoting

*n(n-1) = 15*

n=15 ; n= 16

Another Problem

**P(n,5) = 20 P(n,3)**

I have no idea how to do it

Hoping for replies, thanks - Jan 13th 2009, 02:52 AMGrandadPerms and coms
Hello ImKoAre you sure the first one isn't

**C(n, 2)**? This would give:

$\displaystyle \frac{n(n-1)}{2!} = 15$

$\displaystyle \Rightarrow n(n-1)=30$

$\displaystyle \Rightarrow n^2-n-30=0$

$\displaystyle \Rightarrow n = 6$ or $\displaystyle -5$

But $\displaystyle n > 0$, so the solution is $\displaystyle n = 6$.

For the second question:

$\displaystyle n(n-1)(n-2)(n-3)(n-4) = 20n(n-1)(n-2)$

$\displaystyle \Rightarrow (n-3)(n-4) = 20$

Can you do it now?

Grandad

- Jan 13th 2009, 03:44 AMImKo
For question number 2

How many answers should it have?Should it have 2? or 1 only? - Jan 13th 2009, 03:47 AMIsomorphism
- Jan 13th 2009, 03:50 AMImKo
Wait here is what I got for question number 2

n=23 or n=24 Is that correct?Basing from the solution provided

And what should be the reason for me to remove one of them in order for it to be only one answer? - Jan 13th 2009, 04:05 AMIsomorphism
If we are still talking about "

, then both your answers are wrong.Didnt the problem reduce to solving the quadratic $\displaystyle (n-3)(n-4) = 20 $?(As shown by Grandad...)**P(n,5) = 20 P(n,3)"**

$\displaystyle (n-3)(n-4) = 20 \Leftrightarrow n^2 -7n -8 = 0$. Solve this quadratic...

Always verify your answers by substituting it back in the original equation and verify if your answer is right... - Jan 13th 2009, 04:21 AMImKo
Oh My mistake!! Sorry

so using quadratic formula I got

n = (7 + sqrt17 ) / 2 or n = (7 - sqrt17) / 2

Is that right?

Now which should I eliminate - Jan 13th 2009, 04:29 AMmr fantastic
You should have realised that n is a

**positive integer**therefore:

1. Your solutions to the quadratic $\displaystyle n^2 - 7n - 8$ for n are clearly wrong.

2. The answer to the question*Which of the two correct solutions to the quadratic equation should I eliminate?*is implied by the word in red ..... - Jan 13th 2009, 04:37 AMImKo
Oh! I forgot that the other one is negative, sorry

Is my last solution correct already? - Jan 13th 2009, 04:38 AMGrandadQuadratics
Hello ImKo -

I really think you need to do some work on solving quadratic equations. Your answers using the quadratic formula were incorrect, because you missed a minus sign. But before using the formula, you should always try to factorise - and $\displaystyle n^2-7n-8$ is one of the easiest quadratic expressions to factorise.

Can you do it?

It will look like this: $\displaystyle (n - ??)(n + ??)=0$

This will give you two possible values of $\displaystyle n$, only one of which is appropriate - see mr fantastic's hint.

Grandad

- Jan 13th 2009, 04:45 AMImKo
Sorry I'm a bit careless on solving equations and usually miss the easier approach.. Thanks allot

so the answer would be n = -1 or n = 8

then n = 8 ? - Jan 13th 2009, 04:50 AMGrandadPermutations