View Roling 2 dice what are the chances of P 8,3 C 8,3
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Please explain your question. As written, it does not mean anything.
Originally Posted by Plato Please explain your question. As written, it does not mean anything. I'm trying to use the combinations rule with C8,8. Here is what I've done so far: 8!/8!(8-8)! = 8!/0 = 0? Is this the right way to go about this problem?
Originally Posted by Edbaseball17 I'm trying to use the combinations rule with C8,8. 8!/8!(8-8)! = 8!/0 = 0? I still do not know what this has to do with two dice. However what you have done is incorrect. $\displaystyle (8-8)!=0!=1$, by definition zero factorial equals to 1.
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