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

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Do you know the formula for what C(n,k) is?
n!/((n-r)!r!) n=13 r=4 =715 n=7 r=4 =35 n=6 r=4 =15
n is the pool or what you are choosing from and k is the select or the amount you can take

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Other answers:

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?
Looks correct to me. Now you just simplify that as much as you can.
yeah but when i do i get the wrong answer :[
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?
i thought i could take the compliment... did i do it right?
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.
And that should be a C(6,4) not a C(7,3). Sorry about that...
ohhhh you subtract them? i thought i could combine them by multiplying them. is there anyreason as to why i couldnt do that?
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.
And I've got to go now, feel free to ask more questions as a response, and I'll take a look later tonight.
ok thank you!!!

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