How many ways are there to put 6 b***s in 3 boxes if the b***s are not distinguishable and neither are the boxes?
+118673 If the b***s are all the same and the boxes are all the same then the only difference is how many b***s are in the boxes.
I counted them and I get 7 ways
0,0,6
0,1,5
0,2,4
0,3,3
1,1,4
1,2,3
2,2,2
Since all the b***s and all the boxes are the same, the only difference can be is how many b***s are in the boxes.
+118673 If all the b***s are different and all the boxes are different then
The first ball can go into any of 3 boxes
the second ball can go into any of 3 boxes
.....
so there will be
$${{\mathtt{3}}}^{{\mathtt{6}}} = {\mathtt{729}}$$ options
+118673 If the b***s are all the same and the boxes are all the same then the only difference is how many b***s are in the boxes.
I counted them and I get 7 ways
0,0,6
0,1,5
0,2,4
0,3,3
1,1,4
1,2,3
2,2,2
Since all the b***s and all the boxes are the same, the only difference can be is how many b***s are in the boxes.
+129852 Yep....that answer seems reasonable, Melody......
3 points from me......
+1314 WRONG. im sorry, but WRONG. that would be the coorect answer for this problem though:
How many ways are there to put 6 b***s in 3 boxes if the b***s are distinguishable and the boxes are distinguishable?
now THAT would be 729.
The quickest approach here is to simply list the possibilities. To be more organized, we'll use casework based on the number of b***s in the most full box.
The most full box has 6 b***s: There is only 1 possibility, 6-0-0.
The most full box has 5 b***s: There is only 1 possibility, 5-1-0. (5-0-1 isn't a possibility because the boxes are indistinguishable.)
The most full box has 4 b***s: There are 2 possibilities, 4-2-0 and 4-1-1.
The most full box has 3 b***s: There are 2 possibilities, 3-3-0 and 3-2-1.
The most full box has 2 b***s: There is only 1 possibility, 2-2-2. This is the only possibility because for this case there can be at most 2 b***s in each box.
Therefore there are 7 ways to arrange the b***s.
