Question

A soccer player takes 8 penalty kicks. In how many different ways can he/she make exactly 5?

Answer 1

\[\dbinom{8}{5}=\frac{8\times 7\times6}{3\times 2}\]

Answer 2

assuming it means what i think it means

Answer 3

Could you explain your work satellite?

Answer 4

i am assuming the question is "in how may ways can she make 5 out of 8" meaning for example
S, F, S, S, S, F, S ,F

Answer 5

where S is success and F is failure. so this question is really asking how many ways can you pick 5 slots out of a set of 8, or equivalently 3 slots out of a set of 8. we call this "8 choose 3" written
\[8C3\] some math classes
\[\dbinom{8}{3}\] usually

Answer 6

to compute this there is a formula, which is
\[\dbinom{n}{k}=\frac{n!}{k!(n-k)!}\] but in practice you do not use the formula

Answer 7

we can reason as follows. suppose you want to choose two objects from a set of 8.
you have 8 choices for first slot, 7 for second and 6 for third, so by counting principle this has
\[8\times 7\times 6\] ways to do it, but because you cannot tell for example (abc) apart from (bac) we have overcounted by 3!, the number of ways to arrange 3 items. that is why there is a 3! in the denominator. and that is probably more than you wanted to know

Answer 8

no, keep going, unless you were done, if thats the case than I am a bit confused by the last comment you made

Answer 9

oh ok. no i was done but i can elaborate.

Answer 10

suppose you have 8 friends and you are going to invite 3 to dinner. lets call the friends
A, B, C, D,E, F, G, H

Answer 11

how many ways can you do this? that is, how many ways can you pick 3 out of your 8 friends?

Answer 12

you have 8 choices for the first friend, then 7 choices for the second, and finally 6 choices for the third. so by counting principle you have
\[8\times 7\times 6\] ways to do this

Answer 13

but this is not quite correct, because this counts permutations as different, and they make no difference in this problem. for example, suppose you pick
A, D, F

Answer 14

by out counting method we counted all these as different
A,D,F
A, F, D,
D,A,F
D,F,A
F, A, D,
F, D, A

so for every group of three we picked we counted them each 6 times when we wrote
\[8\times 7\times 6\]

Answer 15

thats is why the number of ways to pick 3 from a set of 8 is not
\[8\times 7\times 6\] but rather it is
\[\frac{8\times 7\times 6}{3\times 2}\]

Answer 16

Alright. It is more clear, and I understand why there needs to be a denominator, but how did you come to that denominator?

Answer 17

2 as in take them or not take them?

Answer 18

well how many ways can you permute three symbols? more generally how many ways can you permute k symbols? that is
\[k!\]

Answer 19

to be honest I've been around permutations awhile but i dont understand it when you say permute, im sorry

Answer 20

oh ok

Answer 21

change there order

Answer 22

okay

Answer 23

for example with two symbols ab i can have
ab
ba
with three
abc
acb
bac
bca
cab
cba

Answer 24

those are abc's permutations?

Answer 25

the last six i mean

Answer 26

yes

Answer 27

What are abc's combinations?

Answer 28

the number of ways to permute k symbols is
\[k!\]

Answer 29

Is that even a valid question?

Answer 30

it is important when you answer these problems to know if "order counts" or not. so for example in my problem about friends going to dinner, order makes no difference. what i was trying to explain was why we overcounted when we wrote
\[8\times 7\times 6\]

Answer 31

if you want, look at pascal's triangle and that will give a good explanation of what
\[\dbinom{n}{k}\] is and how to compute it

Answer 32

http://ptri1.tripod.com/

Answer 33

so in permutations order does count, which is why we cannot use it for that scenario, but in the soccer case, we would use permutations because order counts?

Answer 34

Wait, does order count in the soccer case? Great now I'm confusing myself

Answer 35

Now i see where you got the 3 x 2 from, just 3 factorial. But how do I know if it applies to the soccer case? I'm pretty sure order does not matter in the soccer case so it would be combinations not permutations. therefore the denominator on the soccer kicks is 5 factorial because you are making 5 shots and the numerator is 8 x 7 x 6 x 5 x 4, so the answer is 56, which is what you got! Thanks for your time.