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So given the premise in the title, part (a) asks "How many different ways can you choose 2 marbles of one color and 2 of another color?" and the answer is

${6\choose 2}{7\choose 2}{7\choose 2}$

If I am understanding this correctly, the first comes from the fact that we are "choosing" two colors first and then the ${7\choose 2}{7\choose 2}$ comes from choosing 2 marbles from out of the 7 (for both colors).

However, in the next part it asks "How many different ways can you choose 2 of one color, 1 of a different color, and 1 of yet another color?" and I thought it would be ${6\choose 3}{7\choose 2}{7\choose 1}{7\choose 1}$, but the video I was watching said that the answer was

${6\choose 1}{5\choose 2}{7\choose 2}{7\choose 1}{7\choose 1}$

The video explained this was the answer because the color where we get choose 2 marbles is "distinct," but honestly, I have no clue what they meant by that. Thank you in advanced for your help.

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  • $\begingroup$ You could fix your attempt in the second problem by choosing which of the three colors you picked is the color from which two marbles are drawn. Then your answer would be $$\binom{6}{3}\binom{3}{1}\binom{7}{2}\binom{7}{1}\binom{7}{1} = \binom{6}{1}\binom{5}{2}\binom{7}{2}\binom{7}{1}\binom{7}{1}$$ As Stephen Donovan pointed out, it matters from which color two marbles are drawn. $\endgroup$
    – N. F. Taussig
    Jan 6, 2021 at 9:44

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Basically all they mean is that a situation where you pick $2$ red, $1$ green, and $1$ blue is different from where you pick $2$ green, $1$ red, and $1$ blue. If you just pick the three colors without this distinction, these two situations are counted as if they were the same. This isn't a necessary distinction in the first question because both colors got the same number of marbles, so we can treat them all the same.

The reasoning behind the video's solution is first you pick a color to be the color with $2$ marbles, so pick $1$ out of a possible $6$. Then, pick the other two colors from the remaining $5$. Then you choose the marbles out of each set, giving you the other three factors.

Hope this helps!

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  • $\begingroup$ Hello! This definitely helped to clear things up so thank you! Just to make sure, we use the binomial coefficient when we don't want to count something like "1 blue then 1 green" as different than "1 green then 1 blue," correct? So the reason why we have that ${5\choose 2}$ instead of ${5\choose 1}{4\choose 1}$ is to avoid overcounting? $\endgroup$
    – floriations
    Jan 6, 2021 at 5:42
  • $\begingroup$ That is correct: $\binom{5}{1}\binom{4}{1} = \frac{5!}{(5-2)!}$, which is the function we would use if we were counting pairs of colors where order matters. Here, we're counting where the order doesn't matter so this would be considered overcounting. In fact, we're overcounting by a factor of $2!$, the number of possible orders for each pair, so that's where the factor of $k!$ in the denominator of $\binom{n}{k}$ comes from: it's the factor we overcount by if we count each order separately. $\endgroup$
    – Stephen Donovan
    Jan 6, 2021 at 5:54

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