Thanks jaxis for pointing out the difference.
Here the gifts are different hence there is exist a much more intuitive and easy method as gmat1011 explained.
gmat1011 wrote:Quick question for Rahul: In your stick method ---
If I put 16 different sticks (each standing for a gift) + 3 separators
Then why doesn't (19!)/3!*1! yield the same result... If the gifts are all different then its like dividing by 1! 16 times in the denominator, right? Just as you divide by 16! if the items are all identical. Why doesn't that work here?
Let's analyze for a small number of gifts, say 3 and 2 children.
Say the 4 gifts are A, B, and C and the children are x and y.
Now according to simple application of "stick-separator" method, there are four possible combinations: (0, 3), (1, 2), (2, 1), and (3, 0).
But as the gifts are different, (1, 2) can have three possible sub-arrangements : (A, B + C), (B, A + C), and (C, A + B). Same goes for (2, 1). Hence a total of 8 different combinations.
Whereas if you try to modify the "stick-separator" method as you did, then the number is much more because it takes some false sub-arrangements into account. Such as it interprets (0, 3) as (0, A + B + C), (0, A + C + B), (0, B + A + C), (0, B + C + A), (0, C + A + B), and (0, C + B + A).
Hope this helps to clear the confusion.
