Question

There are 15 members of the show choir. In how many ways can you arrange 4 members in the front row?

Answer 1

So how many different people can we put in the first chair? 15. Now how many in the second chair? Well considering that we've already used one guy, only 14 different ways. Continue down and you can see there are 15*14*13*12 different ways to sit the people in those 4 chairs.

Answer 2

first, take 4 members of 15 members it means 15C4 (i supposed the result is x). so to get 4 members in the front row : xP4
maybe...

Answer 3

so it would be 12 ways?

Answer 4

if im understanding correctly..

Answer 5

im not sure, just still confirm from the other ones...

Answer 6

@Kainui
Suppose you chose:
Alice Bob Chris Dan

Then you tried again and chose:
Bob Alice Chris Dan

Would your method double count these?

Answer 7

Hmmm, actually I think I am misreading the problem, sorry.

Answer 8

@Hero , @hartnn .. please verify that my answer was wrong :)

Answer 9

\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]So: \[
\binom{n}{k} \times k! = \frac{n!}{(n-k)!}
\]This is just the falling factorial from \(n\) to \(n-k\): \[
n \times (n-1) \times (n-2) \times .. \times (n-k)
\]

So I think you guys would get the same solution.

Answer 10

In this case \(n =15\), \(k=4\), and \(n-k = 11\)

Answer 11

Whoops, should have written: \[
\frac{n!}{(n-k)!} = n \times (n-1) \times (n-2) \times \dots (n-k+1)
\]

Answer 12

\(n-k+1=12\)

Answer 13

@RadEn Does what I'm saying make sense?

Answer 14

arrangement means permutation
\(\huge ^{15}P_4\)

Answer 15

so, we neednt the combinatoric, here ?

Answer 16

sorry, @wio . i was still confuse to understanding this problem..

Answer 17

@RadEn since you have selected the 4 using 15C4, you need to arrange them in 4! ways, resulting in 15C4 x 4! or 15P4 ways

Answer 18

yes, the finally.. i got it, thanks for ur confirms.
@wio, hartnn, and sirm3d :)