A Permutation Problem-Urgent

**ImKo** · Jan 13th 2009, 02:43 AM

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

**Grandad** · Jan 13th 2009, 02:52 AM

Hello ImKo

Originally Posted by ImKo

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

Are you sure the first one isn't C(n, 2)? This would give:

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

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

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

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

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

For the second question:

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

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

Can you do it now?

Grandad

**ImKo** · Jan 13th 2009, 03:44 AM

For question number 2

How many answers should it have?Should it have 2? or 1 only?

**Isomorphism** · Jan 13th 2009, 03:47 AM

Originally Posted by ImKo

For question number 2

How many answers should it have?Should it have 2? or 1 only?

It can have two .... however here there is only 1 solution

**ImKo** · Jan 13th 2009, 03:50 AM

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?

**Isomorphism** · Jan 13th 2009, 04:05 AM

Originally Posted by ImKo

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?

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

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

Always verify your answers by substituting it back in the original equation and verify if your answer is right...

**ImKo** · Jan 13th 2009, 04:21 AM

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

**mr fantastic** · Jan 13th 2009, 04:29 AM

Originally Posted by ImKo

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

You should have realised that n is a positive integer therefore:

1. Your solutions to the quadratic $n^2 - 7n - 8$ 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 .....

**ImKo** · Jan 13th 2009, 04:37 AM

Oh! I forgot that the other one is negative, sorry

Is my last solution correct already?

**Grandad** · Jan 13th 2009, 04:38 AM

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 $n^2-7n-8$ is one of the easiest quadratic expressions to factorise.

Can you do it?

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

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

Grandad

**ImKo** · Jan 13th 2009, 04:45 AM

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 ?

**Grandad** · Jan 13th 2009, 04:50 AM

Originally Posted by ImKo

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 ?

Correct!

Grandad

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