有關階乘的難題(分波入盒)

For convenience, we give the name of the boxes as A, B, C and D and then, we analyze the question from the view of how many possible combinations of numbers of balls that can be put into each individual box so as to sum up to a total of 10 (since the balls are identical, we need not considerate further to which ball is put into which box) as follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Aug08/Crazyprob1.jpg


圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Aug08/Crazyprob2.jpg

So summing up, we have 286 different ways to put those 10 balls into the 4 different boxes.

(a1 , a2 , a3 , a4) maps to (a1 , a1+a2+1 , a1+a2+a3+2 , a1+a2+a3+a4+2) = (b1 , b2 , b3 , b4) where 0<=b1<b2<b3<=b4=12. check for bijection and the answer is 12C3

2008-08-12 20:55:41 補充：
A silly mistake... it should be 13C3

2008-08-12 20:55:42 補充：
A silly mistake... it should be 13C3

Suppose there are ten identical balls:
O O O O O O O O O O
And we want to insert 3 'separators' (denoted by X) to separate the balls into 4 groups. The following figure gives one of the examples:
O X O O O O X O O O O O X
(This is equivalent to the bags with 1, 4, 5 and 0 ball(s) respectively.)
Then the problem is equivalent to find the number of ways of choosing 3 objects (3 Xs) from 13 objects (10 Os and 3Xs).
Hence the number of ways
= 13C3
= 13!/[(3!)(10!)]
= 13(12)(11)/6
= 286.
(Remark: Similar arguments may be generalized to m identical balls and n bags, where m and n are any positive integers.)

第一、二個袋空，
第三、四個袋各裝五個。

2 balls in a bag