Agreed. Great analysis.
andreagonzalez2kI work it out a bit backwards (I suppose) but get the same result going through the slow way. Ive always followed this way to keep the calculations and reasoning straight in my head (although I think it adds an extra step):
(1st) for each case I divide the different items into identical “stacks”
(2nd)then I account for any over counting if we have “stacks” of the same size
(3rd)then I arrange each of the “identical” stacks and shuffle them among the 4 different children, treating any child who gets 0 more than once as “identical elements”
Case 1: 5-0-0-0
(1st) no. of ways to choose candies
5 c 5 = 1
(2nd) we can arrange this 1 stack among the 4 children in:
4!/3! Ways (we need to remove the arrangements in which the children who receive 0 are over counted in 4!)
(5 c 5) * (4!/3!) = 1 * 4 = 4
Case 2: 4-1-0-0
(1st) no of ways to break chocolate into “identical stacks”
(5 c 4) * (4 c 1) = 5
And
(2nd) for each of the ways to break down the Items into non-distinct stacks, we can shuffle the stacks among the 4 children:
4!/2! = 12
5 * 12 = 60
Case 3: 3-2-0-0
(1st) choose and select the candies to break into identical “stacks”
(5 c 3) * (2 c 2) = 10
And
(2nd) for each of the 10 ways to break down into non-distinct stacks, we can shuffle around and arrange the identical stacks in:
4!/2! = 4 * 3 = 12
10 * 12 = 120
Case 4: 3-1-1-0
(1st) here is where the way I do it may differ a little (though we end up with the same calculation): 1st, we choose the ways to break down the different chocolates into stacks:
(5 c 3) * (2 c 1) * (1 c 1)
But, for the stacks that are of the SAME SIZE (1 and 1) we will be over counting each one through the selection method - right now, we are ignoring the ordering. So multiply the above result by (1 / 2!)
(5! / 3! 1! 1!) * (1 / 2!) = 10 ways
And
(2nd) for each of these 10 ways to break down the different chocolates into non-distinct stacks, we can then shuffle the stacks among the 4 children in;
4! = 24 ways
10 * 24 = 240
Case 5: 2-2-1-0
(1st) breaking down into identical stacks, but again we have two stacks of the same size (and will be over counting)
(5 c 2) * (3 c 2) * (1 c 1) * (1 / 2!) = 15 ways
And
(2nd) for each of the 15 ways to beak the Choc down into non-distinct stacks, these stacks can be shuffled around the 4 children in:
4! = 24 ways
15 * 24 = 360
Case 6: 2-1-1-1
(1st) (5 c 2) (3 c 1) (2 c 1) (1 c 1) * (1 / 3!) = 10
And
(2nd) 4! = 24
10 * 24 = 240
Adding up the cases:
4 + 60 + 120 + 240 + 360 + 240 =
1024
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