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.

MagnusMathematician Mar 21, 2020

#1**0 **

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

CalTheGreat Mar 21, 2020

#2**-3 **

you can burn all the cherry candy and all the melon candy away and buy the children some chocolate candy

TitaniumRome Mar 21, 2020

#3**0 **

Seriously?

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

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

CalTheGreat
Mar 21, 2020

#5**+1 **

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

MagnusMathematician
Mar 21, 2020

#4**+1 **

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.

MagnusMathematician Mar 21, 2020

#6**+2 **

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!

jfan17 Mar 21, 2020

#7**+1 **

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!

CalTheGreat Mar 21, 2020