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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- anonymous

assuming it means what i think it means

- anonymous

Could you explain your work satellite?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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
\[8\times 7\times 6\]

- anonymous

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
\[\frac{8\times 7\times 6}{3\times 2}\]

- anonymous

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

- anonymous

2 as in take them or not take them?

- anonymous

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

- anonymous

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

- anonymous

oh ok

- anonymous

change there order

- anonymous

okay

- anonymous

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

- anonymous

those are abc's permutations?

- anonymous

the last six i mean

- anonymous

yes

- anonymous

What are abc's combinations?

- anonymous

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

- anonymous

Is that even a valid question?

- anonymous

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
\[8\times 7\times 6\]

- anonymous

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

- anonymous

http://ptri1.tripod.com/

- anonymous

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?

- anonymous

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

- anonymous

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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.