# Permutation question!

• Jun 12th 2013, 03:30 PM
Gurp925
Permutation question!
Hi guys, The question is $7Pn=42$ from here i changed to $\7!(7-n)!=42$ and crossed multiplied giving me the wrong answer the right answer however is two how do we do these?
• Jun 12th 2013, 03:36 PM
topsquark
Re: Permutation question!
Quote:

Originally Posted by Gurp925
Hi guys, The question is $7Pn=42$ from here i changed to $\7!(7-n)!=42$ and crossed multiplied giving me the wrong answer the right answer however is two how do we do these?

It's actually
$_7 P _n = \frac{7!}{(7 - n)!} = 42 \implies 7! = 42 \cdot (7 - n)!$

Or you could reduce
$\frac{7!}{(7 - n)!}$
directly.

-Dan
• Jun 12th 2013, 03:50 PM
Plato
Re: Permutation question!
Quote:

Originally Posted by Gurp925
Hi guys, The question is $7Pn=42$ from here i changed to $\7!(7-n)!=42$ and crossed multiplied giving me the wrong answer the right answer however is two how do we do these?

You might just do it by inspection: $7\cdot 6=42$. So what is $n=~?$
• Jun 12th 2013, 03:59 PM
Gurp925
Re: Permutation question!
love Plato's method answer = 2 yay! and once we get to this portion $7!=42(7-n)!$ how do we get the answer two from here ?
• Jun 12th 2013, 04:16 PM
Plato
Re: Permutation question!
Quote:

Originally Posted by Gurp925
love Plato's method answer = 2 yay! and once we get to this portion $7!=42(7-n)!$ how do we get the answer two from here ?

\begin{align*} 7! &= 42(7-n)!\\6! &=6(7-n)!\\ 5! &=(7-n)!\end{align*}

So $n=~?$
• Jun 12th 2013, 09:36 PM
ibdutt
Re: Permutation question!
Plato has explained very well. However we can also proceed as under
7! = 42 * ( 7-n)!
7*6* 5! = 42 * ( 7-n)!
5! = ( 7-n)!
that gives 7-n=5 OR n = 2
• Jun 13th 2013, 12:55 AM
Gurp925
Re: Permutation question!
Great answers thanks so much guys.