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A committee of $6$ people is to be chosen from $8$ students and $6$ teachers so that it contains at least $3$ students and at least $2$ teachers. What is the number of ways this can be done?

I'm having a lot of trouble with this because normally it is only dealing with $1$ variable: How many ways can $x$ fit in $y$ slots. Whereas this has $2$ variables with $6$ slots.

My take was to do it this way:

$8 \cdot 7 \cdot 6 \cdot 5 = 1800$

and

$6 \cdot 5$

because this takes up $6$ slots.

$8 \cdot 7 \cdot 6$

and

$6 \cdot 5 \cdot 4$

which totals $2286$. Where am I going wrong?

P.S. My English is poor, so do correct me.

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  • $\begingroup$ how many ways are there to select 3 students and 3 teachers ? how may ways are there to select 4 students and 2 teachers ? $\endgroup$ – user451844 Jul 26 '17 at 12:50
  • $\begingroup$ I thought that was what I calculated? $\endgroup$ – Po Chen Liu Jul 26 '17 at 12:54
  • $\begingroup$ there are 8*7*6 ways to choose 3 students and 6*5*4 ways to choose 3 teachers ( assuming you care about order of them being chosen, if not divide each by 3! =6 then you get 8*7*5*4 = 56*20=1120 then we have the possibility of 4 students and 2 teachers and get 7*2*5 *3*5 = 1050 and 1050+1120 =2170 $\endgroup$ – user451844 Jul 26 '17 at 13:06
  • $\begingroup$ thank you for the answer, could you expand on why it's 7*2*5*3*5? $\endgroup$ – Po Chen Liu Jul 26 '17 at 20:02
  • $\begingroup$ for that second part because you divide 8*7*6*5 by 24 = 4! ( the number of ways to arrange 4 things in different orders) and get 7*2*5, then you have 6*5 and you divide it by 2!=2 because that's how many ways you can order the two things) so together you get 7*2*5*3*5=1050 and you can then add to the previous result to get the full result. $\endgroup$ – user451844 Jul 26 '17 at 20:11
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The order of selection does not matter. Therefore, you should be using combinations rather than permutations.

Since the six-person committee must have at least three students and at least two teachers, it either has three students and three teachers or four students and two teachers.

Three students and three teachers: We can select three of eight students and three of the six teachers in $$\binom{8}{3}\binom{6}{3}$$ ways.

Four students and two teachers: We can select four of the eight students and two of the six teachers in

$$\binom{8}{4}\binom{6}{2}$$

ways.

Total: Since the two cases are disjoint, the number of ways we can select a committee of six people from eight students and six teachers that contains at least three students and at least two teacher is

$$\binom{8}{3}\binom{6}{3} + \binom{8}{4}\binom{6}{2}$$

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This what I would do:

Choose 3 among students: $ \binom{8}{3}=\frac{8\times 7 \times 6}{3\times 2\times 1}= 56 ways$

Choose 2 among teachers: $\binom{6}{2}=15 ways $

Choose 1 among the people left unchosen: $\binom{9}{1}=9 ways $

Therefore, there are $56 \times 15 \times 9$ ways

THIS ANSWER IS WRONG! SEE COMMENT BELOW

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  • $\begingroup$ You are counting committees with three teachers and three students three times, once for each of the $\binom{3}{2}$ ways you could reserve two spots for the teachers. To see this, consider the committee $$s_1, s_2, s_3, t_1, t_2, t_3$$ You count this committee in three ways. \begin{align*} & s_1, s_2, s_3, \color{blue}{t_1}, \color{blue}{t_2}, t_3\\ & s_1, s_2, s_3, \color{blue}{t_1}, t_2, \color{blue}{t_3}\\ & s_1, s_2, s_3, t_1, \color{blue}{t_2}, \color{blue}{t_3}\end{align*} By similar argument, you count each committee with four students and two teachers four ways. $\endgroup$ – N. F. Taussig Jul 26 '17 at 19:30
  • $\begingroup$ Your answer is very clear and yes you are right! $\endgroup$ – Chrysanthi Pas Jul 27 '17 at 20:58

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