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How to calculate the odds of 4 specific numbers appearing in a sample of $100$ random numbers (range $1-3000$), knowing that all the $4$ numbers will share a factor (say, $20$) and they will be the only ones in the sample to share it. Order does not matter.

I am confused about two things in particular:

  1. Since I am looking only at the numbers divisible by $20$ in the range, is it relevant to consider the total number of events as $100$ or as $4$? Of course, I assume that I need to take into account the odds of having exactly $4$ $20-$ divisible numbers among $100$ either way.

  2. Since order does not matter, should I divide the probability of getting these four specific numbers by $4$! ?

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    You have an amount of $100$ integers $\in[1,3000]$, where $4$ integers are divisible by $20$ and the rest are not, and you're asking what is the probability of picking these $4$ numbers? Well, the answer is simply $\frac{1}{\binom{100}{4}}$. – 2017-01-16
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    Let me clarify: I do not know in advance that these 4 numbers ARE among the 100 random integers, this is part of the question. Also, does a simple combination function for n=100 take into account the whole range of events? I mean, it is possible (and seems likely) that you will not get any of these four (out of 3000) numbers in the 100 sample of random integers. – 2017-01-16

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I take it that you are using "odds" loosely for probability.
If not, you should be able to convert to odds.

There are $150$ multiples of $20,\;2850$ that are not, in the specified range.

There is just one way to choose the specific $4$, and $\binom{2850}{96}$ to choose the rest,
thus $Pr = \frac{\binom{2850}{96}}{\binom{3000}{100}}\;$ or in fuller form,$\;\frac{\binom44\binom{2850}{96}}{\binom{3000}{100}}$

ADDED

If you actually mean any four multiples of $20$, not specific four,

$Pr = \frac{\binom{150}{4}\binom{2850}{96}}{\binom{3000}{100}}$

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    I am wondering about changes in resulting probability if distribution of numbers in the range is irregular. For example, if I know that out of the total 100 numbers 40 fall in the range 1 - 200 and 3 of the numbers I seek are among them, while the remaining 60 cover the 201 - 3000 range with only 1 number sought among them - can I just treat these as two independent events (1st - 3 numbers out of 1 - 200 among 40, 2nd - 1 number out of 201 - 3000, 60 numbers), using the formula above for each of them separately and just multiply the resulting probabilities? – 2017-09-04
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    Yes, with the stipulations you are making. – 2017-09-05
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    I have a late follow up question. If I assume that repetition is allowed, should I still use the above formula only substituting the binomial coefficient (n k) everywhere with the (n+k-1 k)? When I did, as it seemed logical to do so, the difference of the resulting probability surprised me by being far lower (by few orders of magnitude, c. 1000x), less than I expected. Given that repetition doesn't affect the probability of having four 20-divisible integers in the range but only the number of possible combinations of the 4 that are found such a change seems unjustified by the data. – 2018-04-06