anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
its on permutations and combinations algebra 2
anonymous
  • anonymous
@Preetha
anonymous
  • anonymous
@mathstudent55 ???

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More answers

anonymous
  • anonymous
It is combinations. And the result is same for the all four dinners.
anonymous
  • anonymous
how do i do it?
anonymous
  • anonymous
I can do the problems that are already set up but not word problems where i have to set it up
anonymous
  • anonymous
brb
anonymous
  • anonymous
You 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
  • anonymous
eek i got a weird answer
anonymous
  • anonymous
what did you get?
anonymous
  • anonymous
\[11/18144\]
anonymous
  • anonymous
that was evaluating the right side
anonymous
  • anonymous
ohh 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
  • anonymous
The general formula is: \[\left(\begin{matrix}c \\ n\end{matrix}\right) = \frac{ c! }{ n! \times (c - n)! }\]
anonymous
  • anonymous
i got 220
anonymous
  • anonymous
is that for the first dinner
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok how do i find the others
anonymous
  • anonymous
For 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
  • anonymous
not sure how to solve that?
anonymous
  • anonymous
its 9/3??
anonymous
  • anonymous
Check my answer, I generalized the formula for you
anonymous
  • anonymous
\[\left(\begin{matrix}c \\ n\end{matrix}\right) = \frac{c!}{n! \times (c-n)!}\]
anonymous
  • anonymous
i got 84
anonymous
  • anonymous
yeah, probably, I didn't calculated
anonymous
  • anonymous
ok 3rd?
anonymous
  • anonymous
How many items do we left? We used 6 of them so far right?
anonymous
  • anonymous
6
anonymous
  • anonymous
yes, you will chose 3 from 6. Use the formula that I gave you. Did you understand how to use the formula?
anonymous
  • anonymous
mm let me see ill try
anonymous
  • anonymous
i got 20
anonymous
  • anonymous
yes. I think
anonymous
  • anonymous
ok for the fourth then all of the variables in the formula you gave are 3 right
anonymous
  • anonymous
yes, you have learned ;)
anonymous
  • anonymous
i think so but the problem is that I'm getting 1/6?
anonymous
  • anonymous
you should get 1. Check if you didn't make any calculation mistakes.
anonymous
  • anonymous
or i guess i would just say one because there isn't anything left
anonymous
  • anonymous
yes :)
anonymous
  • anonymous
oh haha okay thank you so much (:
anonymous
  • anonymous
Anytime ;)

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