Piano Club Problem: Finding Hours for Elevens & Twelves

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Discussion Overview

The discussion revolves around a combinatorial problem involving the selection of students from two grades, elevens and twelves, and determining the number of hours associated with each grade based on specific conditions. The focus is on mathematical reasoning and problem-solving techniques.

Discussion Character

  • Mathematical reasoning, Homework-related

Main Points Raised

  • One participant proposes a formula for the number of hours for elevens and twelves based on combinations, but expresses uncertainty about the next steps.
  • Another participant defines the number of grade 11 students as $a$ and grade 12 students as $b=30-a$, providing formulas for selecting pairs from each grade.
  • It is noted that the total number of hours can be expressed as a linear equation involving $a$ based on the combinations calculated.
  • Substitutions are suggested to simplify the equation, leading to a linear equation in $a$.
  • Participants confirm that they arrive at the same value for $a$, which is 12, but the context of this conclusion is not fully explored.

Areas of Agreement / Disagreement

Participants generally agree on the calculations leading to the value of $a$, but there is no explicit discussion of whether this value is the final answer or if further verification is needed.

Contextual Notes

The discussion does not clarify the assumptions behind the problem or the implications of the derived value for $a$. There may be additional constraints or interpretations of the problem that are not fully addressed.

Ilikebugs
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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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