MHB Piano Club Problem: Finding Hours for Elevens & Twelves

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The discussion focuses on solving a problem involving the distribution of hours for grade elevens and twelves in a piano club. It establishes a relationship between the number of grade 11 students (a) and grade 12 students (b), leading to equations for selecting pairs of students from each grade. The key equation derived is 2N1 + 4N2 + 3N3 = 1392, which simplifies to a linear equation in terms of 'a'. After substituting and simplifying, the solution reveals that there are 12 grade 11 students. The conversation emphasizes the importance of careful substitution and algebraic manipulation to reach the solution.
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View attachment 6199 so if a= elevens and b= twelves, the number of hours for elevens are C(a,a-2) times 2, and the number of hours for twelves are C(b,b-2) times 4. I don't know where to go from here
 

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Okay, if we define the number of grade 11 students to be $a$, then there are $b=30-a$ grade 12 students. The number of ways to pick a pair of grade 11 students is:

$$N_1={a \choose 2}=\frac{a(a-1)}{2}$$

The number of ways to pick 2 grade 12 students is:

$$N_2={30-a \choose 2}=\frac{(30-a)(29-a)}{2}$$

And the number of ways to pick a pair consisting of one grade 11 student and 1 grade 12 student is:

$$N_3=a(30-a)$$

And so, using the information given in the problem, we may state:

$$2N_1+4N_2+3N_3=1392$$

Substitute for $N_i$, and you should ultimately obtain a linear equation in $a$. What do you find?
 
I don't know.
 
Ilikebugs said:
I don't know.

I edited your post to translate it into English. Please don't use text-speak abbreviations.

Okay, what I suggested is to substitute as follows:

$$2\left(\frac{a(a-1)}{2}\right)+4\left(\frac{(30-a)(29-a)}{2}\right)+3\left(a(30-a)\right)=1392$$

Distribute:

$$a(a-1)+2(30-a)(29-a)+3a(30-a)=1392$$

Okay, now distribute again, and collect like terms...what do you have?
 
1740−29a=1392

348=29a

12=a?
 
Ilikebugs said:
1740−29a=1392

348=29a

12=a?

Yes, I got $a=12$ as well. (Yes)
 
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