evaluate P(6,2)
You mean permutations ? like 6 P 2 (an nPr) ?
premutations
\(\Large\color{red}{ \rm _6~P~_2 = }\) \(\LARGE\color{blue}{ \rm \frac{6!}{(6-2)!} }\)
\(\LARGE\color{blue}{ \rm = \frac{6!}{4!}=\frac{4! \times 5 \times 6}{4!} }\)
so it would be 4?
Cab you finish the problem from here ?
Can you ... ?
no
Okay, So, do you know what `6!` and `4!` is ?
not really
\(\large\color{red}{ \bf 7! =1\times2\times3\times4\times 5 \times 6 \times7 }\) \(\large\color{red}{ \bf 5! =1\times2\times3\times4\times 5 }\) \(\large\color{red}{ \bf 9! =1\times2\times3\times4\times 5 \times 6 \times7\times8 \times9 }\) this is the meaning behind the exclamation mark here. Get it? Math-wise, \(\large\color{red}{ \bf n! =1 \times 2 \times ~~....~~(n-1) \times n }\)
yeah
So \(\large\color{blue}{ \rm 6!=1\times2\times 3 \times 4 \times 5 \times 6 }\) AND therefore, \(\large\color{blue}{ \rm 6!=(4!) \times 5 \times 6 }\) COrrect ?
coorect
*correct
So, in permutation labeled as ` n P r ` you are choosing r (things) out of n (things). The order matters (as opposed to combinations labeled with ` n C r ` ) Saying that 1,2,3,4 would NOT be the same as 1,3,2,4 when it comes to permutations. The formula is \(\LARGE\color{blue}{ \rm nPr= \frac{n!}{(n-r)!} }\)
In your case you have \(\LARGE\color{blue}{ \rm 6~P ~2 }\)
So, \(\LARGE\color{blue}{ \rm 6~P~2= \frac{6!}{(6-2)!} }\)
Now you just need to simplify it, \(\LARGE\color{blue}{ \rm 6~P~2= \frac{6!}{(6-2)!} }\) \(\LARGE\color{blue}{ \rm 6~P~2= \frac{6!}{4!} }\) Lets re-write ` 6! ` on the top. \(\LARGE\color{blue}{ \rm 6~P~2= \frac{4! \times 5 \times 6}{4!} }\) \(\LARGE\color{blue}{ \rm 6~P~2= \frac{4! \times 30}{4!} }\) ` 4! ` cancel. \(\LARGE\color{blue}{ \rm 6~P~2= 30 }\)
Makes sense ? (tell me when you go over what I typed ...) :)
yes makes since thank you so much! :)n
i actually know what im doing now thanks
Join our real-time social learning platform and learn together with your friends!