voodoochild wrote:
EvaJager wrote:
Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box.
For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box.
Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes.
For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls.
For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities.
Total of 90 + 60 = 150 possibilities.
Answer D.
Eva,
I was able to compute the combinations for distributing balls in 2-2-1 and 3-1-1 way. However, I used 3! for the arrangement of the boxes. Let's say that the boxes are labeled X Y and Z -- and we are using (2,2,1) type of placement of balls. Wouldn't the order of boxes matter? i.e. 3!? That is, I could have any of the three boxes in the first spot, 2 for the second spot and 1 for the third one?
Example : abcde are balls and three boxes are xyz =>
X=[AB] Y=[cd] Z=[e]
X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6
I am a bit confused. Still thinking....Please help me. thanks
You keep the boxes fixed, you permute the balls.
For example, if you have in the boxes |AB| |CD| |E|. This is obtained from ABCDE.
You also have |CD| |E| |AB|, which is obtained from CDEAB, an arrangement already included in the 5! permutations of the balls.
For any permutation of the balls, in the 2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities:
|AB| |CD| |E|
|AB| |C| |DE|
|A| |BC| |DE|
So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example:
AB in the first box is the same as BA, and CD in the second box is the same as DC.
All the arrangements
|AB| |CD| |E|
|BA| |CD| |E|
|AB| |DC| |E|
|BA| |DC| |E|
should be counted just once.
Therefore, 5!*3/(2! * 2! * 1!).
Thanks for your help. I think that I see your point. Do you think that if the boxes are placed on a rack in a store (order of the three boxes will matter -correct?), then we will have to multiply by 3!? That is, for 2-2-1 case, 5C2*3C2*1*(3!) ---Please let me know your thoughts....
Here's how I approached this problem :- for 2-2-1 case, the combinations will be 5C2*3C2*1C1 (choose 2 balls from 5 balls - then - choose 2 balls from the remaining 3 balls - then - choose 1 ball from 1 ball)
Now, since the boxes are labeled differently, we need to permute them. I initially thought of multiplying by 3! because of these combinations :
........ditto for 3-1-1 case.
I am a bit confused.