+54 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.
+2094 We can't see the number of pieces of cherry candy and melon candy. If you tell me. I'd be happy to help!
+2973 you can burn all the cherry candy and all the melon candy away and buy the children some chocolate candy
+2094 Seriously?
Anyway, to find out how many candies the kids can get, do
((#of melon candy)! + (#of cherry candy)!)/4!.
+54 I tried your way, but it wasn't correct. Thanks for trying though.
+54 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.
+499 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!
+2094 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!
