What's probability to choose the 6 correct numbers in a lottery (made up of 42 numbers)? I do know how to do it with combinatorics $${42 \choose 6} = (42·41·40·39·38·37)\div ( 1·2·3·4·5·6) = 502450786$$
but how can you get the answer reasoning fully explained?
First of all, you computed the number of sets of $6$ numbers of $42$. Each of these outcomes is equally probable, and hence the probability of you picking exactly the one that's drawn is therefore:
$$\frac{1}{{42 \choose 6}}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}{42 \cdot 41 \cdot 40 \cdot 39 \cdot 38 \cdot 37}$$
You can also get this as follows. You need to have the first number that's drawn, and since you picked $6$ numbers, the probability of that is $\frac{6}{42}$. Ok, but after you get the first one correct, you also need the second one correct, and since you have $5$ numbers left, and there are still $41$ numbers left that can be chosen, the probability of that is $\frac{5}{41}$. But now you likewise also need the third number correct, the fourth, fifth, and sixth, and so the probability to get them all is:
$$\frac{6}{42} \cdot \frac{5}{41} \cdot \frac{4}{40} \cdot \frac{3}{39} \cdot \frac{2}{38} \cdot \frac{1}{37}$$
... which is of course the same as the above.