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The problem goes as follows:

A committee of $7$ is to be formed from $9$ boys and $4$ girls. In how many ways can this be done when the committee consists of

1. At least $3$ girls?

2. At most $3$ girls?

The way that I'm solving the first part of this problem is by first finding number of ways to choose $3$ girls, which is $^4C_3$. Then I add the one remaining girl to the number of boys and choose $4$ more committee members, which is $^{10}C_4$. The total ways to form the committee, hence, is

$${^4}C_3\times {^{10}}C_4 = 840$$

I solve the second part in a similar fashion. Both my answers seem to be wrong. I have understood how the right answer is to be obtained, but can't figure out why the method above yields the wrong answer. Can somebody explain it to me?

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    $\begingroup$ You have overcounted the case where four girls are picked. Girl'A' might have been picked with the other girls or she might have been picked with the selection of boys. Use addition principle instead. How many ways can it be done with exactly three girls? How many ways can it be done with exactly four girls? Add the results. $\endgroup$ – JMoravitz Jun 22 '16 at 14:08
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A committee of $7$ is to be formed from $9$ boys and $4$ girls. In how many ways can this be done if the committee includes at least $3$ girls?

To select a committee of seven with at least three girls, we must select four boys and three girls or three boys and four girls. That means selecting four of the nine boys and three of the four girls or selecting three of the nine boys and all four girls, which can be done in $$\binom{9}{4}\binom{4}{3} + \binom{9}{3}\binom{4}{4}$$ ways.

Where did you go wrong?

Your groups of four people and ten people are not distinct since you are selecting from $14$ of the $13$ available people. Consequently, you counted each committee that contains four girls four times, once for each way you could have designated three of the four girls to be the three selected girls or, equivalently, once for each of the four ways you could have designated one of the four selected girls to be among the other ten people available to serve on the committee. Notice that $$\binom{9}{4}\binom{4}{3} + \binom{4}{3}\binom{9}{3}\binom{4}{4} = \binom{4}{3}\binom{10}{4}$$

A committee of $7$ is to be formed from $9$ boys and $4$ girls. In how many can be done if the committee includes at most $3$ girls?

You can add the cases in which zero, one, two, or three girls are selected.

$$\binom{9}{7}\binom{4}{0} + \binom{9}{6}\binom{4}{1} + \binom{9}{5}\binom{4}{2} + \binom{9}{4}\binom{4}{3}$$

However, it is easier to count the complementary case that all four girls are selected and subtract that from the total number of ways of selecting a committee of seven people from the thirteen available people.

$$\binom{13}{7} - \binom{9}{3}\binom{4}{4}$$

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