Question

There are 15 members of the show choir. In how many ways can you arrange 4 members in the front row? Permutation or combination? Solve. @johnweldon1993

Answer 1

In permutations, the order of items matters. In combination, order doesn't matter. Since here, order does matter, you'd do 15P4 or P(15,4), equivalent to 15*14*13*12.

Answer 2

What. Ok. Are you sure it is Permutations?

Answer 3

Yes. If the four people in the front row are called Bob, Bill, Stacy, and Emily, and they were arranged that way, it would be different than Emily, Bob, Stacy, and Bill.

Answer 4

How come it's not combinations?

Answer 5

In combinations, the arrangement of items does not matter.

Answer 6

So Bob, Bill, Stacy, and Emily would be the same as Emily, Bob, Stacy, and Bill if it were combinations.

Answer 7

But, the question asks how many different ways can they be arranged, which tells us that it;s a permutation.

Answer 8

I believe the answer is 32760, that is 15*14*13*12

Answer 9

Number of ways = 15 P 4 = 32760 ways

Answer 10

The above is true

Answer 11

@david111 I don;'t want the answer. I need step by step.

Answer 12

15*14*13*12

Answer 13

@johnweldon1993 Would it be like this??\[_{15}P _{4}\]

Answer 14

So you're selecting 4 from 15 to be arranged in the front row. That's 15P4, which is equal to 15*14*13*12, which gives you 32760.

Answer 15

Indeed @YanaSidlinskiy

Everything that has been said so far has been true :)

Answer 16

Ok. Tell me if im right. Ok?

Answer 17

Go ahead.

Answer 18

Wait... I'm kind of stuck... This is what I've got so far. \[_{15}P _{4} = \frac{ 15! }{ (15-4) }\] Don't know what to do next...

Answer 19

It's 15!/(15-4)!, and n! = n*(n-1)*(n-2) and so on until *1. So for example, 6! would be 6*5*4*3*2*1, and 7! would be 7*6*5*4*3*2*1. So here you have 15!/11!.

Answer 20

\[\frac{ 15! }{ 11! }\]

Answer 21

What? I don't get this>_<

Answer 22

Well now you evaluate

look at it like this

\[\large \frac{15!}{(15 - 4)!} = \frac{15 \times 14 \times 13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2}{11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}\]

Now look that after 11....they all just cancel out...so we just have

\[\large \frac{15!}{(15 - 4)!} = \frac{15 \times 14 \times 13\times12\cancel{\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2}}{\cancel{11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}}\]

So all we have is

\[\large 15 \times 14 \times 13 \times 12 = \space?\]

Answer 23

And since 15! can divide by 11! because 15! is 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1, you can just cancel out the 11*10*9*8*7*6*5*4*3*2*1 so you have a numerator of 15*14*13*12 and a denominator of 1 because the 11! has canceled out.

Answer 24

Lol a repeat of john's answer.

Answer 25

That's what I was looking for @johnweldon1993 That's what I actually needed!:) Thanx:) For everyone who helped!:)

Answer 26

Kudos to Mr. John and you're welcome. :)

Answer 27

Haha you did all the work @aurorablue01 so enjoy the earned medal...

and anytime @YanaSidlinskiy :)

Answer 28

Lol, thanks :)

Answer 29

Now, I've gotta go do me some math homework, so see you later, peeps!

Answer 30

The order doesn't matter it's combination