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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    \[\dbinom{8}{5}=\frac{8\times 7\times6}{3\times 2}\]

  2. anonymous
    • 5 years ago
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    assuming it means what i think it means

  3. anonymous
    • 5 years ago
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    Could you explain your work satellite?

  4. anonymous
    • 5 years ago
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    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

  5. anonymous
    • 5 years ago
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    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

  6. anonymous
    • 5 years ago
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    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

  7. anonymous
    • 5 years ago
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    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

  8. anonymous
    • 5 years ago
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    no, keep going, unless you were done, if thats the case than I am a bit confused by the last comment you made

  9. anonymous
    • 5 years ago
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    oh ok. no i was done but i can elaborate.

  10. anonymous
    • 5 years ago
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    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

  11. anonymous
    • 5 years ago
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    how many ways can you do this? that is, how many ways can you pick 3 out of your 8 friends?

  12. anonymous
    • 5 years ago
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    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

  13. anonymous
    • 5 years ago
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    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

  14. anonymous
    • 5 years ago
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    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 \[8\times 7\times 6\]

  15. anonymous
    • 5 years ago
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    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 \[\frac{8\times 7\times 6}{3\times 2}\]

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

  17. anonymous
    • 5 years ago
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    2 as in take them or not take them?

  18. anonymous
    • 5 years ago
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    well how many ways can you permute three symbols? more generally how many ways can you permute k symbols? that is \[k!\]

  19. anonymous
    • 5 years ago
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    to be honest I've been around permutations awhile but i dont understand it when you say permute, im sorry

  20. anonymous
    • 5 years ago
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    oh ok

  21. anonymous
    • 5 years ago
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    change there order

  22. anonymous
    • 5 years ago
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    okay

  23. anonymous
    • 5 years ago
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    for example with two symbols ab i can have ab ba with three abc acb bac bca cab cba

  24. anonymous
    • 5 years ago
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    those are abc's permutations?

  25. anonymous
    • 5 years ago
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    the last six i mean

  26. anonymous
    • 5 years ago
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    yes

  27. anonymous
    • 5 years ago
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    What are abc's combinations?

  28. anonymous
    • 5 years ago
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    the number of ways to permute k symbols is \[k!\]

  29. anonymous
    • 5 years ago
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    Is that even a valid question?

  30. anonymous
    • 5 years ago
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    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 \[8\times 7\times 6\]

  31. anonymous
    • 5 years ago
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    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

  32. anonymous
    • 5 years ago
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    http://ptri1.tripod.com/

  33. anonymous
    • 5 years ago
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    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?

  34. anonymous
    • 5 years ago
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    Wait, does order count in the soccer case? Great now I'm confusing myself

  35. anonymous
    • 5 years ago
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    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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