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A class consists of $15$ boys of whom $5$ are prefects. How many committees of 8 can be formed if each consists of at least $2$ prefects?

I had the solution following solution in mind but that seems to be incorrect. can anyone please explain to me why?

First, I select $2$ students from the perfect category using $\binom{5}{2}$ and then I select remaining $6$ students combined from both the perfect and unperfect category i.e. $\binom{13}{6}$. So the answer becomes $\binom{5}{2}\times \binom{13}{6} = 17160$. But, the correct answer is $5790$. Can you please help me?

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  • $\begingroup$ How many committees...? Only one committee can be formed at any time, because you'll need at least 16 individuals to form two committees of 8 each, and you have 15 only. Unless you're asking how many ways there are to form a committee.... $\endgroup$
    – CiaPan
    Dec 4, 2016 at 17:49

4 Answers 4

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The solutions proposed by @JSchoone and @THELONEWOLF are correct.

You are counting committees with more than two prefects more than once.

There are $$\binom{5}{k}\binom{10}{8 - k}$$ ways to choose a committee with exactly $k$ prefects and $8 - k$ non-prefects. Adding the results for $2 \leq k \leq 5$ gives you the correct solution proposed by THE LONE WOLF. JSchoone's method of subtracting the number of committees with fewer than two prefects from the total is computationally more efficient.

In your method, when you choose two prefects and six more committee members from the remaining thirteen boys, your method counts each committee with three prefects $\binom{3}{2}$ ways, once for each of the three ways you could have chosen two of the three selected prefects as the two designated prefects. Your method counts each committee with four prefects $\binom{4}{2}$ ways, once for each of the six ways you could have chosen two of the four selected prefects as the two designated prefects. Your method counts the committee with all five prefects $\binom{5}{2}$ ways, once for each of the ten ways you could have designated two of the five selected prefects as the two designated prefects. Notice that $$\binom{5}{2}\binom{10}{6} + \binom{3}{2}\binom{5}{3}\binom{10}{5} + \binom{4}{2}\binom{5}{4}\binom{10}{4} + \binom{5}{2}\binom{5}{5}\binom{10}{3} = 17160$$

Also, notice that you are selecting the committee of eight boys from $5 + 13 = 18$ of the fifteen boys, while JSchoone and THE LONE WOLF are choosing the committee from the five prefects and non-prefects, that is, from the $10 + 5$ available boys. You ran into trouble because your group of five boys and group of thirteen boys are not separate groups.

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    $\begingroup$ okay. got it. thank you so much.! you're a genius. $\endgroup$ Dec 4, 2016 at 11:30
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There should be minimum $2$ perfects in the committee, there is no limit for maximum of perfects. So, The number of ways can be counted as:

$1$. $2$ perfect and $6$ non perfect..... Number of ways become ${{5}\choose{2}}\times{{10}\choose{6}}$.

$2$. $3$ perfect and $5$ non perfect....... Number of ways become ${{5}\choose{3}}\times{{10}\choose{5}}$.

$3$. $4$ perfect and $4$ non perfect.....Number of ways become ${{5}\choose{4}}\times{{10}\choose{4}}$.

$4$. $5$ perfect and $3$ non perfect.....Number of ways become ${{5}\choose{5}}\times{{10}\choose{3}}$.

Add all the ways and get the answer.

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The total number of committees of $8$ is $\binom{15}{8}$. The number of committees of $8$ with no prefects is equal to $\binom{10}{8}$. The number of committees with exactly one prefect is equal to $\binom{5}{1} \cdot \binom{10}{7}$.

Then the answer to your question is $\binom{15}{8} - \binom{10}{8} - \binom{5}{1} \cdot \binom{10}{7}=6435 - 45 - 5\cdot 120 = 5790.$

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15C8-(10C8 + 10C7*5C1) = 5790

At least two means, total number of ways - number of ways to make committee without any prefect + number of ways to make committee with one prefect and 7 non prefects.

15C8 - total number of ways to choose 8 people committee. 10C8 - number of ways not to choose any prefects. 10C7*5C1 - number of ways to choose one prefect and 7 non prefects

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    $\begingroup$ Welcome to MSE. Your answer adds nothing new to the already existing answers. $\endgroup$ Jan 3, 2019 at 10:06

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