Question

Three balls are to be placed in three different boxes, not necessarily one ball in each box. Any box can hold one, two or three balls. Find the number of ways the balls can be placed
(a) if they are of the same colour and therefore indistinguishable,
(b) if they are all of different colours.

Answer 1

a)
first case: first box has no balls. thus the other two can have (1,2),(2,1)or(3,0),(0,3),4 combinations.
second case: first box has one ball. thus the other two can have(1,1),(0,2),(2,0), 3 combinations.
third case: first box has two balls. thus the other two can have (0,1),(1,0) ,2 combinations.
forth case: first box has three balls. thus the other two only have (0,0) 1 combination.

Adding them up and since the first box can be any box(they are different),
3X(4+3+2+1)=30 combinations.

Answer 2

in (b), you'll have to consider their respective permutations
such as in the first case,(1,2) is the same as 3C1 and (2,1) is the same as 3C2.

do you need further help?

Answer 3

thanks..for (a) the answer is 10..i had listed down all the possiblities n iv got the answer..but is there any other way to find the answer?
and for (b) the answer is 27..

Answer 4

so i guess there wasn't a need to times 3 then.
yeah....i don't think so==this question is not really classical.

Answer 5

ok..thanks..

Answer 6

treating the balls as o's the use 2 dividers to make 3 boxes

ie ooo||

now rearrange thses

there are 5 objects so 5! but you have 3 and 2 that are the same so divide by 3! and 2!

5!/(3!2!)=(5*4*3*2*1)/(3*2*1*2*1)=10

or just \(_5C_3=10\)

Answer 7

thanx zarkon..iv understood now:) can u please help me for part (b)???

Answer 8

3 choices for 1st ball
3 choices for 2nd ball
3 choices for 3rd ball
3*3*3=27