In how many of ways can 5 balls be placed in 4 tins if any

@[Bunuel]
I am not getting the answer for this question
In this question if we can also consider following cases :

(1 CASE) Tin 1 (ALL 5 BALLS) + TIN 2 ( O BALLS)+ TIN 3 ( O BALLS) + TIN 4 ( O BALLS) --- TOTAL PERM-S.: 4!/3! .

(2 CASE) Tin 1 ( 4 BALLS) + TIN 2 ( 1 BALLS)+ TIN 3 ( O BALLS) + TIN 4 ( O BALLS) --- TOTAL PERM-S.: 4!/2! .

(3 CASE) Tin 1 ( 3 BALLS) + TIN 2 ( 2 BALLS)+ TIN 3 ( O BALLS) + TIN 4 ( O BALLS) --- TOTAL PERM-S.: 4!/2! .

(4 CASE) Tin 1 ( 3 BALLS) + TIN 2 ( 1 BALLS)+ TIN 3 ( 1 BALLS) + TIN 4 ( O BALLS) --- TOTAL PERM-S.: 4!/2!

(5 CASE) Tin 1 ( 2 BALLS) + TIN 2 ( 1 BALLS)+ TIN 3 ( 1 BALLS) + TIN 4 ( 1 BALLS) --- TOTAL PERM-S.: 4!/3!


ADDING 1+2+3+4+5 = 4+12+12+12+4 =44 <<< WHATS WRONG IN THIS APPROACH ??

From what I understand of a previous question, the problem with this approach is that it treats all the balls as equal. If all the balls are equivalent then
your approach is correct (I think) but imagine every ball was a different colour, there are suddenly a lot more combinations than 44.

There is another similar question here about putting letters in a box, if they are different letters then your approach doesn't work, but if they are equivalent (like junk mail) then your approach does work.

Maybe look that up.

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