Piano club hours problem with combinations

[Ilikebugs]

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

 

[MarkFL]

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?

 

[Ilikebugs]

I don't know.

 

[MarkFL]

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?

 

[Ilikebugs]

1740−29a=1392

348=29a

12=a?

 

[MarkFL]

1740−29a=1392

348=29a

12=a?

Yes, I got $a=12$ as well. (Yes)