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# I'm stuck on this question

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I have 8 identical pieces of cherry candy and 7 identical pieces of melon candy. Find the number of ways I can distribute this candy to 4 children.

Mar 21, 2020
edited by MagnusMathematician  Mar 21, 2020

#1
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We can't see the number of pieces of cherry candy and melon candy. If you tell me. I'd be happy to help!

Mar 21, 2020
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you can burn all the cherry candy and all the melon candy away and buy the children some chocolate candy

Mar 21, 2020
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Seriously?

Anyway, to find out how many candies the kids can get, do

((#of melon candy)! + (#of cherry candy)!)/4!.

CalTheGreat  Mar 21, 2020
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I tried your way, but it wasn't correct. Thanks for trying though.

MagnusMathematician  Mar 21, 2020
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Not a bad idea TR. But rather than burn it why don't you just store it for them ( maybe in your stomach)?

Melody  Mar 21, 2020
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whoops im so sorry I have 8 identical pieces of cherry candy and 7 identical pieces of melon candy. Find the number of ways I can distribute this candy to 4 children.

Mar 21, 2020
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Hey nice to see you again Magnus! This is another case of stars and bars like I was explaining to you earlier. Because we have 8 cherry candies, let's first look at how to distribute that to 4 children. That would be:

\({8+4-1 \choose 8}\)\({11 \choose 8}\)= 165

Now that we have the number of ways to distribute the cherry candies, we can multiply that with the number of ways to distribute the melon candies(correct me if I'm wrong with this reasoning).

Distributing 7 things to 4 people, we get:

\({7+4-1 \choose 7}\)\({10 \choose 7}\)= 120

165 * 120 = 19800 ways. Now that I think about it, I feel like my number is a bit high, but feel free to double check this reasoning!

Mar 21, 2020
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Combinations!! That's it!!! Nice, jfan17.

Just in case you don't know what a combination is, the formula is n!/r!(n-r)!

Hope this helped!

Mar 21, 2020
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Thank you so much!

Mar 21, 2020