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A bag contains 40 balls from 4 different colors: red, white, black and green. There are 10 balls numbered (1, 2, 3, ..., 10) from each color. 5 balls are randomly selected from the bag.

In how many ways can 5 balls be chosen so that no more then 2 balls are of the same color?

Can you help me with this? I know that there are 2 cases, 1: 2 balls with same color & 3 balls with different colors 2: 2 balls with same color, another 2 balls with same color and 1 ball with different color. However, I could not understand my instructor's explanation.

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    $\begingroup$ Splitting into the two cases you described is good. Can you share your attempts at counting the number of outcomes for each case? $\endgroup$ Commented Nov 12, 2020 at 23:55
  • $\begingroup$ 1st case: (4,1) (10,2) (3, 3) (10, 1) (10,1), (10, 1). 2nd Case: (4,3) (10, 2) (10, 2) (10, 1) x3. You can consider the parentheses as combination. 2nd case is multiplied with 3. This is my instructor's solution. 1st case is fine, but 2nd? Why not (4,1) (10, 2) (3, 1) (10,2) (2,1) (10,1) $\endgroup$ Commented Nov 13, 2020 at 0:15
  • $\begingroup$ Note that ${4\choose3}={4\choose1}$, and ${3\choose1}=3$, so the only difference between your answer and your instructor's is that you have ${2\choose1}$ and your instructor doesn't. $\endgroup$ Commented Nov 13, 2020 at 0:22
  • $\begingroup$ thank you, makes sense. i should divide it to two due to repetition. $\endgroup$ Commented Nov 13, 2020 at 0:30

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Responding to comment:

The reason why your attempt is incorrect is because you are double counting. For instance,

"black 1, black 2, red 4, red 3, and green 7"

and

"red 4, red 3, black 1, black 2, and green 7"

are counted separately in your method, due to the $\binom{4}{1} \binom{3}{1}$ computation. If you divide your answer by $2$ to account for this, you match your instructor's solution.

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