Question

Medal given to best answer
I don't know to do combinations or factorial
I have 3 pieces of candy to place in 4 lunch boxes. In how many ways can I do this if:
(d) The candies are all different and the lunch boxes are all the same?

(e) Exactly two of the candies are the same (but the third is different) and all of the lunch boxes are different?

Answer 1

@bibby
love your avatar
getting anything?

Answer 2

is this a challenge problem

Answer 3

Allofus?

Answer 4

Yes its a challenge problem

Answer 5

a. each candy has 4 choices, so 4^3 = 64

b. using the stars and bars formula, (3+4-1)C3 = 6c3 = 20

c. in 1 box , in 2 boxes or 3 boxes: 3c3 +3c2 + 3c1 = 7

d. in 1 box, 2 boxes or 3 boxes = 3

e. in 1 box (4) , in 2 boxes : 2combos (AA|B & AB|A) distributed to boxes in 4!/2! ways, and 1 each a box distributed to boxes in 4!/2! ways = 4+24+12 = 40 ways

Answer 6

@IM5er101
I got
(a)There are 64 ways if everything is different, because C1(candy 1) has four choices of lunchbox, C2 has four choices, and C3 has four choices. 4*4*4 = 64
(b)There are 3 choices because either one lunchbox has all three candies, or one lunchbox has 2 and one has 1, or three lunchboxes each have 1.
(c)There are 20 ways since:
1) each of the 4 lunchboxes could have three candies, (4 possibilities) or
2) one of the 4 has two candies and one of the other 3 could have one candy (4*3 = 12 possibilities) or
3) three of the 4 each have one candy.(C(4,3) = 4 possibilities)
so there are 4 + 12 + 4 = 20 possibilities

Answer 7

I just saw this question on yahoo. saw these sites.

https://answers.yahoo.com/question/index?qid=20130729090147AAam0jy

https://answers.yahoo.com/question/index?qid=20111229101015AADvzIp

Answer 8

Oh the questions are in different order thx!!!