Question

We have 8 pieces of strawberry candy and 7 pieces of pineapple candy. In how many ways can we distribute this candy to 4 kids?

Answer 1

8 * 7 * 4

Answer 2

the website says it's wrong

Answer 3

really ?

Answer 4

yah.

Answer 5

anyone wanna take over

Answer 6

I've tried 56, 224, 1365...

Answer 7

I've been working on this problem for forever XD

Answer 8

@cwrw238

Answer 9

@satellite73

Answer 10

@Loser66

Answer 11

25 minutes I've been waiting. 25 MINUTES!!!

Answer 12

1365

Answer 13

@jdoe0001

Answer 14

How many different ways can we distribute candy to the `first kid`?
We could give him,
`1 strawberry`
`7 pineapple`

orrrrr,
`2 strawberry`
`6 pineapple`

and on and on and on.
I think that gives us \(\large 8!\cdot 7!\) combinations.

But then we have ... 4 kids.. hmm :\
I dunno probability :c sorry...

Seems like it would be a really really big number.
Maybe I'm not interpreting the problem correctly though.

Answer 15

Combinations without replacement.
\(C(n,k)=\binom{n}{k}=\dfrac{n!}{k!(n-k)!}\)

Answer 16

Actually, I think it is two of those... one for each flavor, then they are multiplied.

Answer 17

But it has been a while since I did statistics. Hehe.

Answer 18

hey, so my answer is right, right?

Answer 19

nope. *sigh*

Answer 20

Do you have unlimited guesses? D:
Wanna just try another? 2450.
or maybeeee, 1,411,200.

My calculations are so far apart, you can tell I know nothing about this XD lolol

Answer 21

This is tricky and really bugging me. It isn't just a combination, since you aren't picking 4 pieces or picking 7 pieces, you are distributing 7 items and distributing 8 items to 4 "slots".

I'm thinking that it would work better to think of the kids as the options for where each of the pieces of candy gets distributed. E.g, for the 8 strawberry candies, there are 4 possibilities for the first piece, 4 for the second piece, etc, until you reach 4 for the 8th piece.... so that would mean \[4^8=65,536\] options. BUT of course, that isn't accounting for duplicate distributions, e.g., kid 1 gets pieces 1-4 and kid 2 gets 5-8 is no different than kid 2 gets 1-4 and kid 2 gets 5-8.... and I can't quite figure out how to count and factor out those duplicates!??

Answer 22

Use dots and dividers for both the strawberry and pineapples.
For example, if there are 8 candy and 4 people, that means that there need to be 4-1=3 dividers. We just need to put these dividers on the people.
11! ways to arrange everything, 8! ways to arrange the candy (indistinguishable), and 3! ways to arrange the dividers.
So you get

Answer 23

... sorry it ended for some reason
So you get 11!/8!3! = 165.
Do the same with the pinapples.
Final answer (if you are lazy) is 19800.