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