Let
1=a ball is put in a certain box by a certain person.
0=no ball is put in a box by a certain person.
If the way the balls are put in the boxes is random (50/50 chances to appear a 0 or a 1) then all possible variants have the same probability of appearance.
In the case of this problem the total number of combinations possible is:
Z=2^(7*144) (1)
Lets consider now that those 7 people put (or do not put) the balls in the first box first [respecting the order:the first person-having 0 or 1 as possibilities......the seventh person] then they move to the second box and so on until the 144-th box.
THE FIRST BOX
The total number of variants for the first box T [of the form ABCDEFG where A is what the first person put in the first box-a 0 or a 1,B is what the second person put and so on till G-what the seventh person put] is:
T=2^7 (2)
where we have:
7[C]0=7!/{(7-0)!*0!}=1 is the number of variants (of the form ABCDEFG) where in the first box there are 0 balls.
7[C]1= 7 is the number of variants (of the form ABCDEFG) where in the first box there is 1 ball.
7[C]2=21 is the number of variants (of the form ABCDEFG) where in the first box there are 2 balls.
7[C]3=35 is the number of variants (of the form ABCDEFG) where in the first box there are 3 balls.
7[C]4=35 is the number of variants (of the form ABCDEFG) where in the first box there are 4 balls.
7[C]5=21 is the number of variants (of the form ABCDEFG) where in the first box there are 5 balls.
7[C]6=7 is the number of variants (of the form ABCDEFG) where in the first box there are 6 balls.
7[C]7=1 is the number of variants (of the form ABCDEFG) where in the first box there are 7 balls.
The above possibilities can be found easier using 'Pascal''s triangle' rule.The same considerations as above are valid for all boxes till the 144-th.
Next we can calculate the number of combinations of the form:
N=N[1-x1,2-x2,.......,144-x144] ;0 <= x1 <= 7;......0 <= x144 <= 7;
N being the number of distinct variants having x1 balls in the first box,x2 balls in the second box,...,x144 balls in the 144-th box.
N=[1-x1,2-x2,.......,144-x144] ={7[C]x1}*{7[C]x2}*....*7[C]x144 (3)
where 7[C]xk=7!/{[7-xk]!*xk!}
For:
x1=x2=......=x144=3 --->
N[1-3,...,144-3]=144[C]144*{7[C]3}^144=N[144,3]
x1=x2...=x143=3;x144=0 or 1 or 2 or 4,...,7 --->
N[1-3,...,143-3,x144]=144[C]143*{7[C]3}^143*[7^(144-143)]=N[143,3]
144[C]143=144 takes into account the permutations possible.
x1=x2...=x142=3;x143=01,2,4...7,x144=0,1,2,4,...,7 --->
N[1-3,...,142-3,x143,x144]=144[C]142*{7[C]3}^142*[7^(144-142)]=N[142,3]
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.
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x1=3;x2=1,2,5...7;........;x143=01,2,4...7;x144=0,1,2,4,...,7 --->
N[1-3,x2,...,x144]=144[C]1*{7[C]3}^1*[7^(144-1)]=N[1,3]
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The number of distinct variants for which there are at least 3 balls in a single box-N[3]-is the sum of the above terms.In the same manner we can calculate N[4] but we must eliminate the variants where x=3 already taken into account.-->N'[4].
In this case the number of distinct variants for which boxes have 3 or 4 balls is N=N[3]+N'[4] (5).
At most we can calculate the following media for the average number of boxes having 3,4 balls:
M={N[144,3]*144+N[143,3]*143+..+N[1,3]*1+N'[144,4]*144+..+N'[1,4]*1}/{N[144,3]+..+N[1,3]+N'[144,4]+..+N'[1,4]}
Very difficult to make it in practice...
