Question

There are 15 members of the show choir. In how many ways can you arrange 4 members in the front row when order does not matter?

Answer 1

If there are 4 available seats then the first seat could be 1/15 people, the second could be 1/14 the third 1/13 the fourth 1/12 and the fifth 1/11. So for the number of possible combinations you simple multiply 15x14x13x12x11.

Answer 2

360360

Answer 3

right

Answer 4

Actually I gave you the formula for permutations where order would matter. the formula for order not mattering reduces the number of possible combinations considerably.

The formula for this is \[n!/(r!(n-r)!)\]

So in this case

15!/4!(11!)

Answer 5

I think u should multiply 15x14x13x12=32760

Answer 6

That method would give you all possible permutations, however the problem does not want the same 4 kids just arranged in different order because that would be the same "front row" therefore you have you use the equation above to eliminate the excess combinations of the same four kids.

Answer 7

here some guy did this before.
I he the link help you
http://openstudy.com/study#/updates/54a44337e4b0b8a54fb09fab

Answer 8

hope no he
lol

Answer 9

That link says literally the same equation I did.

Answer 10

sorry to be rude but the guy explains it better

Answer 11

No offense, I'm a n00b here. In school to be a teacher. I probably suck at explaining things at this point.

Answer 12

lol it cool

Answer 13

so the answer 32760

Answer 14

or wat

Answer 15

the thing u send me I don't understand

Answer 16

32760 or wat

Answer 17

it is right

Answer 18

No

So in the equation n!/(r!(n−r)!)
n= the total number of kids
r= the number that you are selecting for the front row
I'm assuming they taught you about factorials, denoted by the exclamation point !
So 15! = 15x14x13x12x11x10x9x8x7x6x5x4x3x2x1

use a calculator to get the result for the operation 15!/(4!(11!))
and you will come up with 1365

Answer 19

so the answer is 1365

Answer 20

@Prosper2win That is not correct. That is the possible permutations where the order does matter. However, in this case, order does not matter. Essentially if kids A, B, C, and D are in the front row in a different order D,C,B,A it is still the same kids in the front row. Because we're looking for combinations where order doesn't matter you use the formula above to eliminate excess possibilities.

Yes the answer is 1365. Hopefully you can see why you have to use that particular equation though. Here is another explanation

http://www.mathsisfun.com/combinatorics/combinations-permutations.html

Answer 21

yeah thank u

Answer 22

oh I see thanks