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We have $7$ balls, three red, two white and two blue. How many ways can we select three of them such that at least two are red? So, my answer was: If the balls are identical, then there are $3$ ways: $RRR$, $RRB$ and $RRW$. If the balls are distinguishable, then there are ${7}\choose{3} $=$35$ total possibilities, but there are ${4}\choose{3}$=$4$ ways that have no red balls, and $2 \cdot$${4}\choose{2}$=$12$ with only one ball so there are 19 with 2 red balls or more. But the answer given is 13. Could anyone explain this to me?

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    $\begingroup$ There are $3\cdot\binom{4}{2} = 18$, not $12$, ways with exactly one red ball. $\endgroup$
    – rogerl
    Oct 11, 2021 at 21:02

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You are correct that there are $\binom{7}{3}$ ways to select three of the seven balls and that there are $\binom{4}{3}$ ways to select none of the red balls. However, there are $\binom{3}{1}\binom{4}{2}$ ways to select exactly one of the three red balls and two of the remaining four balls, which yields $$\binom{7}{3} - \binom{4}{3} - \binom{3}{1}\binom{4}{2} = 13$$ ways to select at least two red balls.

Alternatively, we can select exactly two red balls in $\binom{3}{2}\binom{4}{1}$ ways since we must select two of the three red balls and one of the other four balls, and we can select all three red balls in $\binom{3}{3}$ ways. Since these cases are mutually exclusive, we obtain $$\binom{3}{2}\binom{4}{1} + \binom{3}{3} = 13$$ selections with at least two red balls.

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  • $\begingroup$ Why is this reasoning wrong? := There are $\binom{3}{2} $ ways to choose the two read balls. Then there are $\binom{5}{1}$ ways to choose 1 of the remaining balls. Together they make $\binom{3}{2} \times \binom{5}{1} =15 $. $\endgroup$ Mar 25 at 14:01
  • $\begingroup$ @user2277550 Suppose we number the three red balls $r_1, r_2, r_3$. Your method counts the case in which all three red balls are selected three times, once for each of the three ways you could designate one of the three red balls as the remaining ball: $(\{r_1, r_2\}, \{r_3\})$, $(\{r_1, r_3\}, \{r_2\})$, $(\{r_2, r_3\}, \{r_1\})$. That accounts for the two extra selections in your answer. $\endgroup$ Mar 25 at 16:09

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