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nishathomp

  • 2 years ago

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

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

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

  3. nishathomp
    • 2 years ago
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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

  4. KingGeorge
    • 2 years ago
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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?

  5. nishathomp
    • 2 years ago
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    (715)-(525)/(715)

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

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

  8. nishathomp
    • 2 years ago
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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?

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

  10. KingGeorge
    • 2 years ago
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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.

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

  12. nishathomp
    • 2 years ago
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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?

  13. KingGeorge
    • 2 years ago
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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.

  14. KingGeorge
    • 2 years ago
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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.

  15. nishathomp
    • 2 years ago
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    ok thank you!!!

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