# Permutation question!

• Jun 12th 2013, 02:30 PM
Gurp925
Permutation question!
Hi guys, The question is $\displaystyle 7Pn=42$ from here i changed to $\displaystyle \7!(7-n)!=42$ and crossed multiplied giving me the wrong answer the right answer however is two how do we do these?
• Jun 12th 2013, 02:36 PM
topsquark
Re: Permutation question!
Quote:

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

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

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

-Dan
• Jun 12th 2013, 02:50 PM
Plato
Re: Permutation question!
Quote:

Originally Posted by Gurp925
Hi guys, The question is $\displaystyle 7Pn=42$ from here i changed to $\displaystyle \7!(7-n)!=42$ 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: $\displaystyle 7\cdot 6=42$. So what is $\displaystyle n=~?$
• Jun 12th 2013, 02:59 PM
Gurp925
Re: Permutation question!
love Plato's method answer = 2 yay! and once we get to this portion $\displaystyle 7!=42(7-n)!$ how do we get the answer two from here ?
• Jun 12th 2013, 03:16 PM
Plato
Re: Permutation question!
Quote:

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

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

So $\displaystyle n=~?$
• Jun 12th 2013, 08:36 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 12th 2013, 11:55 PM
Gurp925
Re: Permutation question!
Great answers thanks so much guys.