In how many ways can 5 different candies be distributed in 4 identical baskets?	Pick one.
A) 120
B) 51
C) 24
D) 5^4
E) 4^5
5-N-N-N : 5C5 = 1 - OK   4-1-N-N : 5C4 *1C1 = 5 - OK   3-2-N-N : 5C3*2C2 = 10 - OK   3-1-1-N : 5C3*2C1*1C1 = 20 - NO - only 10, once you decide to split the remaining two after choosing the 3 before, no factor of 2 needed   2-2-1-N : 5C2*3C2*1C1 = 30 - NO - it is 15 = 30/2; baskets being identical, doesn't matter which group of 2 you choose first   2-1-1-1 : 5C2*3*2*1 = 60 - NO - only 10, because you only choose those 2 to be placed together, all the other three you put in different baskets, no choices   Total of   1 + 5 + 10 + 10 + 15 + 10= 
<gadget id="calculator">  1 + 5 + 10 + 10 + 15 + 10</gadget>
<output>51</output>
51.   Answer 51.
<result>B</result>
In how many ways can 5 different candies be distributed in 4 identical baskets?