How do I find the probability of picking a science major and an engineering major? 
I was trying to answer part (b) like this:
There are $30 + 24 + 20 + 17 = 91$ students in total.
$\Rightarrow P(\text{science major AND engineering major}) = \dfrac{\frac{20!}{1!(20-1)!}}{\frac{91!}{1!(91-1)!}}\cdot\dfrac{\frac{30!}{1!(30-1)!}}{\frac{91!}{1!(91-1)!}}=\dfrac{20}{91}\cdot\dfrac{30}{91}=\dfrac{600}{8281}$
But this is the wrong answer. Is my method incorrect?
 A: The computation of
$$\frac{20 \times 30}{91 \times 91} \tag1$$
is wrong for two reasons:

*

*In the denominator, once the first person is selected ($91$ choices) there are only $90$ people left for the second choice.  So, the denominator should be changed to $(91 \times 90)$.


*In the numerator, the computation represents the presumption that order of selection is irrelevant.  Such a presumption is okay, as long as the presumption is consistently applied in the denominator.
There are then two ways of accommodating the second bullet point above, both of which lead to the same computation.   It should either be
$$\frac{\binom{20}{1} \times \binom{30}{1}}{\binom{91}{2}} = \frac{20 \times 30}{\binom{91}{2}} \tag2 $$
or
$$\frac{2 \times \left[20 \times 30\right]}{91 \times 90}. \tag3 $$
In (3) above, the denominator presumes that order of selection is to be regarded as relevant.  This explains the leading factor $2$ in the numerator of (3) above, which presents that you could :

*

*pick an engineering major and then pick a science major 
or

*pick a science major and then pick an engineering major.

