# Question about combinations: What is the probability that 3 numbers chosen from 40 are even?

We are choosing from a set of consecutive integers: {1, 2, ..., 40}. We would like to know the probability of choosing 3 even numbers out of 5 total choices. I figured out that it is: $${20 \choose 3} {20 \choose 2}\over {40 \choose 5}$$ I can see why this is correct, (choose 3 from even numbers, choose two from odd numbers, and then the cartesian product of two sets produces all combinations) but for some reason, I can't intuitively understand why the following would be incorrect (the entire fraction below would replace the numerator above): $$20 * 19 * 18 *20 * 19\over {5!}$$ My reasoning here is similar to the first method, except we instead divide by 5! That is, how many ways the 3 even and 2 odd numbers can be arranged, because we are ignoring order. It ends up being off by a factor of 10. Where is my reasoning going wrong, and does the second equation calculate the probability of some other event? If so, what would it be?

Your second calculates the chance you draw three evens and then two odds. You miss all the other orders of odd and even numbers. As there are $${5 \choose 3}=10$$ ways to choose the places the even numbers come in the sequence, your second calculation is low by a factor $$10$$.
• @GrahamKemp: it replaced the numerator, so should be divided by $40 \choose 5$ to get a probability Commented Feb 4, 2020 at 1:47