Ready?..
25 identical sweets must be shared bewteen 1 boy & 1 girl. Each of the children MUST recieve at least 10 sweets each. all sweets must be distrubuted. order not important although I will say this...
I am only looking the total number of unique arrangments, e,g..
1 2 3 4 5
2 4 5 3 1
3 5 3 2 4
the above are displaying exactly the same imformation but in different order such combinations are NOT permited here.
Excuse my lack of mathematical terminologies, (still learning) and merry xmas to you all!
Seasoning Greetings!
So if these 'identical' sweets are in fact distinguishable, there are this many ways:
$${25 \choose 10} + {25 \choose 11} + {25 \choose 12} + {25 \choose 13} + {25 \choose 14} + {25 \choose 15}$$
Added: To try and cut through the conversation in the thread below, let me say this is the number of ways that 25 objects can divided up into two sets, one labelled $B$ for boy, the other labelled $G$ for girl, such that each set has at least 10 members. The calculation here by its construction does not allow for ordering of elements of $B$ and $G$; for example, these two partitions are counted once, not twice:
$$1: B = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, G = \{ 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25 \}$$ $$2: B = \{ 10, 2, 3, 4, 1, 6, 7, 8, 9, 5\}, G = \{ 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25 \}$$