Find the number of ways in which 7 distinct objects can be put in three identical boxes so that no box remains empty.
This is similar to a problem i had in an old course where you had to find integers x,y,z s.t. x + y + z = some +ve integer.
So i think this one can be reworded as the 3 boxes being x,y and z. And n being 7 distinct objects.
Since you want +ve integers what to do is set x' = x-1 etc and then you get x' + y' + z' = 4.
This is done by using the binomial expression where k is the number of 'boxes'
So we get = 15.
Yeah i think my way was if there was 7 identical objects... 15 is waaaay to low.
But how did you get 322? I just did a 'brute force' method and came up about 99 but didnt really check it.
Since each box must contain 1 object it can be simplified down to how many ways can 4 objects be distributed between 3 boxes... This questions bugging me now...
EDIT: Is the answer just ?
Well,this is how it goes.
Since the boxes are identical the problem reduces to the fact distinct objects have to be divided(not distributed) into groups as arrangement among the boxes are immaterial since the boxes are identical
The groupings can be done as follows 1,3,3),(1,2,4),(2,2,3),(1,1,5)" alt="1,3,3),(1,2,4),(2,2,3),(1,1,5)" />
This can be done in following ways:
=
Either the answer is 301 or I have missed a grouping which is not likely
301 is correct.
The number is also , where S is the Stirling Number of the 2nd Kind.
(1,2,4) is same as (1,4,2) since boxes are identical.There is no such thing as the first box or the second or the third .
If 2 groups contain same number of objects then it is required that we divide by 2!.
For example if {a,b,c,d,e,f,g} are the 7 obects.One grouping containing (2,2,3) objects can be described as follows.
{a,b},{c,d}{e,f,g} and another can be {c,d},{a,b}{e,f,g}.
Now arent these groupings same.