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?
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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 $10$.
answered Feb 4, 2020 at 1:15
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$\begingroup$ The value is much greater than 1 and so is not any sort of probability at all. $\endgroup$ Commented Feb 4, 2020 at 1:46
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1$\begingroup$ @GrahamKemp: it replaced the numerator, so should be divided by $40 \choose 5$ to get a probability $\endgroup$ Commented Feb 4, 2020 at 1:47