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anonymous
 one year ago
MEDAL FAN AND TESTIMONIAL:on a camping trip you bring 12 items for 4 dinners for each dinner you use 3 items in how many ways can you choose the 3 items for the first dinner? for the second? for the third? for the fourth?
anonymous
 one year ago
MEDAL FAN AND TESTIMONIAL:on a camping trip you bring 12 items for 4 dinners for each dinner you use 3 items in how many ways can you choose the 3 items for the first dinner? for the second? for the third? for the fourth?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its on permutations and combinations algebra 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is combinations. And the result is same for the all four dinners.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can do the problems that are already set up but not word problems where i have to set it up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You have four dinner. And for all of them, you need to make a decision to pick 3 items among 12 items. So 3 combinations among 12 item is: \[\left(\begin{matrix}12 \\ 3\end{matrix}\right) = \frac{ 12 \times 11 \times 10 }{ 9! \times 3! }\] I had a mistake though, in my first statement, where I said that they were equal. It is not. If you pick 3 items among 12, you will have 9 items left, since you took 3 of them.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0eek i got a weird answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that was evaluating the right side

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh sorry, I forgot to add factorial sign to the 12. It is like this: \[\frac{ 12! }{ 9! \times 3! }\] it is originally this way.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The general formula is: \[\left(\begin{matrix}c \\ n\end{matrix}\right) = \frac{ c! }{ n! \times (c  n)! }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that for the first dinner

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok how do i find the others

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For the second dinner you will calculate this: \[\left(\begin{matrix}9 \\ 3\end{matrix}\right)\] Since we took the first three items, 9 left in the item list.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not sure how to solve that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Check my answer, I generalized the formula for you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left(\begin{matrix}c \\ n\end{matrix}\right) = \frac{c!}{n! \times (cn)!}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, probably, I didn't calculated

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How many items do we left? We used 6 of them so far right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, you will chose 3 from 6. Use the formula that I gave you. Did you understand how to use the formula?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0mm let me see ill try

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok for the fourth then all of the variables in the formula you gave are 3 right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, you have learned ;)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think so but the problem is that I'm getting 1/6?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you should get 1. Check if you didn't make any calculation mistakes.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or i guess i would just say one because there isn't anything left

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh haha okay thank you so much (:
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