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nishathomp

Help with Combinations!!!!!! (C(13,4) - (C(7,4)*C(6,4)))/(C(13,4)

  • one year ago
  • one year ago

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  1. KingGeorge
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    Do you know the formula for what C(n,k) is?

    • one year ago
  2. dreadslicer
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    n!/((n-r)!r!) n=13 r=4 =715 n=7 r=4 =35 n=6 r=4 =15

    • one year ago
  3. nishathomp
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    n is the pool or what you are choosing from and k is the select or the amount you can take

    • one year ago
  4. KingGeorge
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    Basically that's what it it is. But as dreadslicer pointed out above, the formula for C(n,k) is\[C(n,k)=\frac{n!}{k!(n-k)!}\]Using this, you can individually calculate the different values you need. Since you need to find \[\frac{C(13,4)-C(7,4)\cdot C(7,3)}{C(13,4)}\]You need only to find C(13,4), C(7,4), and C(7,3). Can you list these out for me?

    • one year ago
  5. nishathomp
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    (715)-(525)/(715)

    • one year ago
  6. KingGeorge
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    Looks correct to me. Now you just simplify that as much as you can.

    • one year ago
  7. nishathomp
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    yeah but when i do i get the wrong answer :[

    • one year ago
  8. nishathomp
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    the whole problem is this : Jean-Luc's starship encounters an enemy starship. After realizing that he had no choice but to get, Jean-Luc orders his weapons ocer to re 4 weapons at the enemy starship. The weapons ocer randomly selects the 4 weapons to re from a total of 6 photon torpedoes and 7 positron torpedoes. What is the probability that officer selected torpedoes of both types?

    • one year ago
  9. nishathomp
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    i thought i could take the compliment... did i do it right?

    • one year ago
  10. KingGeorge
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    You nearly had it. Just instead of multiplying C(7,4) and C(6,4), subtract both of them. So your equation should be \[\frac{C(13,4)-C(7,4)-C(7,3)}{C(13,4)}\]You take out the choices where you choose all positron torpedoes, and then take out the choices where you chose only photon torpedoes. Then you divide by the total.

    • one year ago
  11. KingGeorge
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    And that should be a C(6,4) not a C(7,3). Sorry about that...

    • one year ago
  12. nishathomp
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    ohhhh you subtract them? i thought i could combine them by multiplying them. is there anyreason as to why i couldnt do that?

    • one year ago
  13. KingGeorge
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    Basically, you have two cases. Case 1, all positron torpedoes. Case 2, all photon torpedoes. When you have two distinct cases like this, you have to add/subtract.

    • one year ago
  14. KingGeorge
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    And I've got to go now, feel free to ask more questions as a response, and I'll take a look later tonight.

    • one year ago
  15. nishathomp
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    ok thank you!!!

    • one year ago
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