A Permutation Problem-Urgent

• Jan 13th 2009, 02:43 AM
ImKo
A Permutation Problem-Urgent
We are to solve for n

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 AM
Perms and coms
Hello ImKo
Quote:

Originally Posted by ImKo
We are to solve for n

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:

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

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

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

$\displaystyle \Rightarrow n = 6$ 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?

• Jan 13th 2009, 03:44 AM
ImKo
For question number 2

How many answers should it have?Should it have 2? or 1 only?
• Jan 13th 2009, 03:47 AM
Isomorphism
Quote:

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 :)
• Jan 13th 2009, 03:50 AM
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?
• Jan 13th 2009, 04:05 AM
Isomorphism
Quote:

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 $\displaystyle (n-3)(n-4) = 20$?(As shown by Grandad...)

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

• Jan 13th 2009, 04:21 AM
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
• Jan 13th 2009, 04:29 AM
mr fantastic
Quote:

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 $\displaystyle n^2 - 7n - 8$ 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 AM
ImKo
Oh! I forgot that the other one is negative, sorry

Is my last solution correct already?
• 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 $\displaystyle n^2-7n-8$ 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.

• Jan 13th 2009, 04:45 AM
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 ?
• Jan 13th 2009, 04:50 AM
Permutations
Quote:

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!