Choosing representatives in class

Total possible combinations = 10C2 = 45.

Exactly 30 contains one or more female => the rest 15 combinations has exactly 2 males each. Therefore, nC2 = 15 => n(n-1) = 30 => n = 6.

Firstly we want to work out how many possible combinations of 2 students out of 10 there are: $10C2=45$ combinations.

Now we can work out the probability of both students being boys.

$Let\quad x\quad be\quad the\quad number\quad of\quad boys\quad in\quad the\quad class.\\ P(x=2)=\frac { 45-30 }{ 45 } =\frac { 15 }{ 45 } =\frac { 1 }{ 3 }$

The probability of the first selection being a boy is $\frac { number\quad of\quad boys\quad in\quad class }{ total\quad number\quad of\quad students\quad in\quad class }$ .

Once we have selected one boy, the number of boys/students in the class available to be selected second have both been reduced by 1 (as one boy has already been picked), so the probability of selecting a boy the second time becomes $\frac { number\quad of\quad boys\quad in\quad class\quad -\quad 1 }{ total\quad number\quad of\quad students\quad in\quad class\quad -\quad 1 }$ .

$\therefore$ we know that $P(x=2)=\frac { x }{ 10 } \times \frac { x-1 }{ 9 } \\ \therefore \quad \frac { x }{ 10 } \times \frac { x-1 }{ 9 } =\frac { 1 }{ 3 }$

Now we can solve this equation to find $x$ , the number of boys in the class.

$\frac { 1 }{ 3 } =\frac { x }{ 10 } \times \frac { x-1 }{ 9 } =\frac { x\times (x-1) }{ 90 } =\frac { { x }^{ 2 }-x }{ 90 }$

Multiply both sides by 90: $30={ x }^{ 2 }-x\\ { x }^{ 2 }-x+30=0$

Factorise: $(x-6)(x+5)=0\\ \therefore \quad x=6\quad OR\quad x=-5$

The number of boys in the class cannot be negative, so $x\neq -5.$

$x=6$ must be the solution, so the number of boys in the class is 6.

Let $x$ as the number of female students. Counting all possible ways to choose one or more female from 10 students :
$xC1$ x $(10-x)C1 + xC2 = 30$
$x$ x $(10-x) + \frac{x(x-1)}{2} = 30$
Solving this equation, give us $x = 4$
Thus, the number of males students are $(10-x) = \boxed{6}$

$m=$ number of males, $f=$ number of females

we have the following system of equations:

$10=m+f$

$30 = \binom{f}{2} + \binom{f}{1} \binom{m}{1}$

solving gives $f=4, m=6$

The possible combinations that equal thirty are 1 Male (M) and 1 Female (F) or 2 Female(F). Therefore: MC1*FC1 + FC2 = 30

Note: M = 10-F

FC2 is: $\frac{F(F-1)}{2}$

This is results in: $(10-F)F + \frac{F(F-1)}{2} = 30$

simplifying/factorizing this equation we get: $(F-4)(F-15)$

Therefore, F = 4 since, F = 15 is not possible M = 10-4 = 6 which is the answer

10C2=45............45-30=15...............nC2=15 on solving gives n=6

there are two cases possible 1) 1 girl selected 2) both are selected x(10-x)2+(10-x)(9-x)=60
in which x is equal to no of girls so ans is 6

Let boys be x. Selecting at least 2 girls -: xC0.10-xC2 + xC1.10-xC1 = 30. x=6

10c2 - (10-x)c2=30 solve it . the logic beind this is that total cases - those in which only boys are selected.here x is the no. of girls