1. 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?

2. Re: Permutation question!

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

3. Re: Permutation question!

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 $n=~?$

4. 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 ?

5. Re: Permutation question!

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 $n=~?$

6. 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

7. Re: Permutation question!

Great answers thanks so much guys.