## 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?

1. anonymous

its on permutations and combinations algebra 2

2. anonymous

@Preetha

3. anonymous

@mathstudent55 ???

4. anonymous

It is combinations. And the result is same for the all four dinners.

5. anonymous

how do i do it?

6. anonymous

I can do the problems that are already set up but not word problems where i have to set it up

7. anonymous

brb

8. 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.

9. anonymous

eek i got a weird answer

10. anonymous

what did you get?

11. anonymous

$11/18144$

12. anonymous

that was evaluating the right side

13. 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.

14. anonymous

The general formula is: $\left(\begin{matrix}c \\ n\end{matrix}\right) = \frac{ c! }{ n! \times (c - n)! }$

15. anonymous

i got 220

16. anonymous

is that for the first dinner

17. anonymous

yes

18. anonymous

ok how do i find the others

19. 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.

20. anonymous

not sure how to solve that?

21. anonymous

its 9/3??

22. anonymous

Check my answer, I generalized the formula for you

23. anonymous

$\left(\begin{matrix}c \\ n\end{matrix}\right) = \frac{c!}{n! \times (c-n)!}$

24. anonymous

i got 84

25. anonymous

yeah, probably, I didn't calculated

26. anonymous

ok 3rd?

27. anonymous

How many items do we left? We used 6 of them so far right?

28. anonymous

6

29. anonymous

yes, you will chose 3 from 6. Use the formula that I gave you. Did you understand how to use the formula?

30. anonymous

mm let me see ill try

31. anonymous

i got 20

32. anonymous

yes. I think

33. anonymous

ok for the fourth then all of the variables in the formula you gave are 3 right

34. anonymous

yes, you have learned ;)

35. anonymous

i think so but the problem is that I'm getting 1/6?

36. anonymous

you should get 1. Check if you didn't make any calculation mistakes.

37. anonymous

or i guess i would just say one because there isn't anything left

38. anonymous

yes :)

39. anonymous

oh haha okay thank you so much (:

40. anonymous

Anytime ;)