Need to understand problems with approach on an "at least" combinations problem

Question

$11$ students are to be chosen from $2$ classes of $20$ students each. At least $5$ need to be chosen from each. Would you please tell me what is wrong with this approach?

First we select $5$ students from each class. The number of ways of doing so $= C(20,5) \times C(20,5)$. Now, $30$ students are left and $1$ needs to be chosen from them $= C(30,1)$. Therefore total number of ways = $(C(20,5))^2 \times C(30, 1)$

The answer is $2(20C5 \times 20C6)$ using cases approach (correct answer). Although the solution for the same is all over the Internet, I don't think there is justification for why my approach is wrong other than simply the fact that my answer is wrong.

You have determined the six students from one class by specifying a particular five first. — Ethan Bolker, Feb 24 '18 at 20:20
That's the second correct approach. I want to understand what's wrong with the first one. @EthanBolker — Arjav Garg, Feb 24 '18 at 20:26
You have overcounted by a factor 0f $6$ because your method finds those $6$ students $6$ ways. — Ethan Bolker, Feb 24 '18 at 20:38

score 3 · Answer 1 · answered Feb 24 '18 at 20:34

Let's abbreviate the names of the students as follows: in the first class the students are $A_1, A_2, \ldots, A_{20},$ and in the second class the students are $B_1, B_2, \ldots, B_{20}.$

Now let's see some examples of how your method could select the $11$ students.

First example: The five students selected from the first class are $A_1,A_2,A_3,A_4,A_5$; from the second, we select $B_1,B_2,B_3,B_4,B_5.$ Now from among the $30$ remaining students we select $A_6.$ Final result: $$\{A_1,A_2,A_3,A_4,A_5,A_6,B_1,B_2,B_3,B_4,B_5\}.$$

Second example: The five students selected from the first class are $A_1,A_2,A_3,A_4,A_6$; from the second, we select $B_1,B_2,B_3,B_4,B_5.$ Now from among the $30$ remaining students we select $A_5.$ Final result: $$\{A_1,A_2,A_3,A_4,A_5,A_6,B_1,B_2,B_3,B_4,B_5\}.$$

Your counting method ends up overcounting by a factor of exactly $6,$ because for each actual way to select the students, you have counted six different ways to select the group of six students from one class, depending on which of those six students is chosen last.

score 2 · Accepted Answer · answered Feb 24 '18 at 20:34

You over counted some cases. Call the students $A_{1},\cdots,A_{20},B_{1},\cdots,B_{20}$. Then using your process, there are more than one way to pick $A_{1},\cdots,A_{6},B_{1},\cdots,B_{5}$: one way is to pick $A_{1},\cdots,A_{5},B_{1},\cdots,B_{5}$ in the first step, and pick $A_{6}$ in the second step. Another way is to pick $A_{2},\cdots,A_{6},B_{1},\cdots,B_{5}$ first and then pick $A_{1}$.

Need to understand problems with approach on an "at least" combinations problem

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