P n C question.Please help.

[akhil.mrt]

Q) In how many ways can we put 4 different balls in 3 different boxes when any box can contain any number of balls?

80

Correct Option:- 81

64

63

82

[nikhil.baveja]

the 1st ball can go in any of 3 boxes so 1st ball has 3 choices.
similarly 2nd ball can go in any 3 boxes, so again 3 choices,
so for all the balls its 3*3*3*3 = 81

[tim]

assuming the balls are distinguishable (eg. different colors) then this is correct. if the balls are indistinguishable, this is a much harder problem and one you wouldn't see on the GMAT (try it out though if you're up for the challenge!)..

[kapilnitt]

Hi Tim,

Even if the balls are same, i think answer is not very tough. Here is my approach:

O = ball
I = Kind of a partition.

O,I,O,O,I,O
Now we just need to arrange them = 6!/(3!2!)= 60

Please correct me, if i'm wrong.

Thanks
Kapil

[RonPurewal]

Hi Tim,

Even if the balls are same, i think answer is not very tough. Here is my approach:

O = ball
I = Kind of a partition.

O,I,O,O,I,O
Now we just need to arrange them = 6!/(3!2!)= 60

Please correct me, if i'm wrong.

Thanks
Kapil

i'm not tim, but...

this approach is correct, except for one thing: it should be 6!/(4!2!), not 6!/(3!2!).
so that's just 15, not 60.

it's also not very hard just to list all of them: just make three columns, one for box 1, one for box 2, and one for box 3.
4 0 0
3 1 0
3 0 1
2 2 0
2 1 1
2 0 2
1 3 0
1 2 1
1 1 2
1 0 3
0 4 0
0 3 1
0 2 2
0 1 3
0 4 0
fifteen possibilities.

the "partition" approach is quite creative, though. well done. just watch your threes and fours!

[kapilnitt]

That's right Ron.
This is the same reason i didn't get 50+ in gmat i.e. i've the right idea, right approach, but silly mistakes bring me down.

Anyway thanks for the correction.

[tim]

:)