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What I did was I calculated the options of putting 16 balls in 4 drawers with the condition of every drawer must have an even number of balls. In order to ensure I have an even number of balls in every drawer I inserted the balls in pairs. So I have 8 pairs to insert to 4 drawers. the result is 165 options ( according to the formula ). In order to solve my question can I do 16 choose 4 ( that are all the options ) minus 165 (even number of balls in every drawer) to get the odd number of balls in each drawer that i'm asked?

Husam
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  • Are the drawers distinct ? And is it like throwing a fair $4$ sided die for putting into the drawers ? – true blue anil Feb 17 '23 at 17:12
  • The drawers are not distinct, and it is fair yes. – Husam Feb 17 '23 at 17:20
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    Your answer of $165$ ways to distribute the $16$ balls so that there are an even number of balls in each drawer only makes sense if the drawers are distinct. – N. F. Taussig Feb 17 '23 at 17:25
  • Are you saying it is a case of identical balls into identical bins ? It should then be quite easy. – true blue anil Feb 17 '23 at 17:26
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    You cannot find the number of distributions with an odd number of balls in each drawer by subtracting the number of ways of distributing an even number of balls in each drawer from the total number of distributions since you could have an odd number of balls in two drawers and an even number of balls in the other two drawers. – N. F. Taussig Feb 17 '23 at 17:30
  • Carefully consider the comment of N.F. Taussig and the answer of Mike Earnest. From these responses, you can infer that either all drawers have an even number of balls, all drawers have an odd number of balls, or exactly two drawers have an even number of balls. Then, consider Stars and Bars theory, which is discussed here and here. ...see next comment – user2661923 Feb 17 '23 at 18:07
  • This means (assuming that all of the variables below must be non-negative integers), that the number of solutions to $$x_1 + \cdots + x_4 = 16$$ must equal the sum of the number of solutions represented by the following equations: $$x_1 + \cdots + x_4 = 8$$ and $$x_1 + \cdots + x_4 = 6$$ and $$\binom{4}{2} \times ~\text{# solutions to} ~x_1 + \cdots + x_4 = 7.$$ In the last computation, $~\displaystyle \binom{4}{2}~$ represents the number of ways of selecting two of the $~4~$ bins to receive one ball each. – user2661923 Feb 17 '23 at 18:12
  • "The drawers are not distinct" : My previous two comments were based on the opposite assumption that the drawers are distinct. Note that your computation of $165$ ways of having all $4$ drawers have an even number of balls, represents the number of non-negative integer solutions to $$x_1 + \cdots + x_4 = 8.$$ However, this implies that the computation of $165$ that you discussed in your answer, also represents the contradictory assumption that the $4$ drawers are distinct. – user2661923 Feb 17 '23 at 19:15
  • it's actually my fault they are distinct. Thanks a lot for the explanation though now I know the difference:) – Husam Feb 17 '23 at 23:33

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Your method does not work. You are counting the number of ways to place the balls such that it is not true that every box is even, which means the number of ways where there is at least one odd box. This is not the same as all boxes being odd.

You came up with a clever way to force all of the boxes to be even; simply place the balls in pairs. Similarly, there is a clever way to force all of the boxes to be odd. Start by placing a single ball in each box, and then distribute the remaining twelve balls in six pairs.

Mike Earnest
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