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Context: there are three candies in a basket (5 red, 4 blue, 5 green). How many ways can you choose 2 candies with different color?

Should I use permutation or combination w/ this? I tried using: 14!/(14-2)! = 182 ways; but it's marked as wrong.

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Hint:

You can choose $5\times4$ pairs of candies s.t. one of them is red and one of them is blue.

Of course there are more possibilities.

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  • $\begingroup$ Sorry, but I still don't get it! So by 5 x 4 is it literally 5 times 4 (like 20)? $\endgroup$ – unexpected Dec 10 '15 at 15:06
  • $\begingroup$ Yes. Do you agree that there are $20$ possibilities to choose a pair of candies such that one of them is red and the other is blue? Btw, what do you mean with: "there are three candies in a basket". Maybe I misunderstand something there. Aren't there $14$ candies in total? $\endgroup$ – drhab Dec 10 '15 at 15:20
  • $\begingroup$ Yes I agree. Also I meant 3 groups. So yeah, there are 14 candies. $\endgroup$ – unexpected Dec 10 '15 at 15:26
  • $\begingroup$ Well then: $5\times 4+5\times5+4\times5=65$ possibilities. The first term corresponds with "red and blue" the second with "red and green" and the third with "blue and green". $\endgroup$ – drhab Dec 10 '15 at 15:30

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