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This is the problem: " In a class there are $40$% more girls than boys. If the probability that delegation of two is made up by a boy and a girl is $\frac {1}{2}$, How many students are in the class?."

I don´t know if "a delegation of two" is translated correctly but it think, it basically means like a group of two.

I tried to solve it the following way:

Let´s call $N_s$ the number of students. Because there are boys and girls, $N_s = N_b + N_g$ (number of students = number of boys + number of girls).

Then the total ways you could form a delegation are: $$ \binom {N_s}{2}$$ So I thought that if could subtract the number of ways you could form a group of of two boys, and then the number of ways you could a group of two girls, from the total ways, you could get the number of ways you could form a group by one boy and one girl. So that´s what I did:

$$ \binom {N_s}{2} - \binom {N_b}{2} - \binom{ N_g}{2} = M$$

In the end, you would have: $$ \frac {M}{\binom {N_s}{2}} = \frac{1}{2} $$.

I don´t write the process because I made a mistake, I don´t know if it is in the way of solving it or if in the process, So I wanted to ask if this way of solving the problem is correct and if there is a better way?

I also use the fact that $N_b + 40$%$N_b = N_g$.

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Let there be $b$ boys and $1.4b$ girls. This gives a total of $2.4b$ students. The number of ways to select two students is $\frac 12\cdot 2.4b(2.4b-1)$. The number of ways to select one boy and one girl is $1.4b\cdot b$, so we get $$\frac 12 \cdot \frac 12 \cdot 2.4b(2.4b-1)=1.4b\cdot b\\ 1.44b^2-0.6b=1.4b^2\\0.04b^2-0.6b=0\\b=15$$ So there are $15$ boys and $21$ girls in the class.

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  • $\begingroup$ But is the way I am solving it correct? $\endgroup$ – Vmimi May 4 '18 at 14:29
  • $\begingroup$ Your approach is fine but incomplete. You never incorporate the fact that there are $1.4$ times as many girls as boys. You mention it at the end. I think my approach to $M$ is simpler as $M=1.4b \cdot b$ $\endgroup$ – Ross Millikan May 4 '18 at 14:54
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Let $g$ and $b$ be the numbers of girls and boys, respectively. Then: $$\begin{cases}g=1.4b \\ \frac{{g\choose 1}{b\choose 1}}{{g+b\choose 2}}=\frac12\end{cases} \Rightarrow \begin{cases}g=1.4b \\ \frac{gb}{\frac{(g+b)(g+b-1)}{2}}=\frac12\end{cases} \Rightarrow \begin{cases}g=1.4b \\ 4gb=(g+b)(g+b-1)\end{cases} \Rightarrow \\ 5.6b^2=2.4b(2.4b-1) \Rightarrow 2.4b=0.16b^2 \Rightarrow b=15 \Rightarrow g=21.$$

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