anonymous
  • anonymous
A soccer player takes 8 penalty kicks. In how many different ways can he/she make exactly 5?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\dbinom{8}{5}=\frac{8\times 7\times6}{3\times 2}\]
anonymous
  • anonymous
assuming it means what i think it means
anonymous
  • anonymous
Could you explain your work satellite?

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anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
oh ok. no i was done but i can elaborate.
anonymous
  • 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
  • anonymous
how many ways can you do this? that is, how many ways can you pick 3 out of your 8 friends?
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
2 as in take them or not take them?
anonymous
  • anonymous
well how many ways can you permute three symbols? more generally how many ways can you permute k symbols? that is \[k!\]
anonymous
  • anonymous
to be honest I've been around permutations awhile but i dont understand it when you say permute, im sorry
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
change there order
anonymous
  • anonymous
okay
anonymous
  • anonymous
for example with two symbols ab i can have ab ba with three abc acb bac bca cab cba
anonymous
  • anonymous
those are abc's permutations?
anonymous
  • anonymous
the last six i mean
anonymous
  • anonymous
yes
anonymous
  • anonymous
What are abc's combinations?
anonymous
  • anonymous
the number of ways to permute k symbols is \[k!\]
anonymous
  • anonymous
Is that even a valid question?
anonymous
  • 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
  • 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
  • anonymous
http://ptri1.tripod.com/
anonymous
  • 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
  • anonymous
Wait, does order count in the soccer case? Great now I'm confusing myself
anonymous
  • 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.

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