probability of picking 4 numbers out of 100 I have $100$ balls and on each ball is labelled a number between $1$ and $100$. No two balls can have the same number, so the interval $1\ldots 100$ is represented by the balls.
I now pick 20 balls at random, one by one without putting them back at any point. I am trying to find the probably of getting 4 certain numbers. The sequence of these 4 numbers does not matter.
Here is my solution: The first time I pick up a ball the probably is $1/100$. The next time it is $1/99$, etc.. So the total probably of getting the four desired numbers is $1/(100\cdot 99\cdot 98\cdot 97)$.
Is this correct?
 A: You have computed the probability that you get your target numbers, in the first $4$ choices, in a certain specified order. That is much smaller than the required probability.
There are $\binom{100}{20}$ equally likely ways to choose $20$ balls.
We find the number of ways to choose $20$ balls that include our $4$ target balls. So we must choose $16$ balls from the remaining $96$ to keep our $4$ target balls company. This can be done in $\binom{96}{16}$ ways.
Thus the required probability is $\dfrac{\binom{96}{16}}{\binom{100}{20}}$. This expression can be greatly simplified. 
A: Your solution is not correct.  But, here's a hint on a different way to proceed:
Hint: You want to count all permutations (of length $20$) of distinct elements in $\{1,2,\ldots,100\}$, such that four fixed numbers (say $1,2,3,4$) are all contained in the permutation.
To do this: pick the $16$ numbers OTHER than the ones you are interested in which will be included; then, pick ANY arrangement of the set of $20$ numbers consisting of the four you want and the new sixteen.
This is the total number of sequences with the desired property; divide by the total number of sequences to get the desired result.
