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We have a class of 100 students and we want to form 4 subgroups of 40,30,20 and 10 students each one. What's the number of different ways to form these subgroups:

So given that we have a partition we have:

$\dbinom{100}{40}\dbinom{60}{30}\dbinom{30}{20}\dbinom{10}{10} $

I'm pretty sure that's O.K. But now the problem goes far as:

Suppose there are 4 friends. Which is the probability that they all end in the same group?

So I'm trying to look it like this: as they are mutually exclusive events, try to add the prob. of the 4 friends ending in the 1st group + ending in the 2nd group and so on... like this:


But given the result is so small I'm pretty sure I'm wrong with my guess. So I'm looking now from this approach: Given that I now for sure 4 friends belong to the first group I don't have 100 available to group 40 but just 96 and to group 36 and so on, like this:


But these approximation seems a little over complex. So, it's my approach fine or what am I missing?


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up vote 1 down vote accepted

I would proceed like this. What is the probability that the four friends are assigned to the big group? Call them A,B,C,D and assume that A is assigned first, then B, then C, then D, then the rest of the $100$ people.

The probability A is assigned to the big group is $\frac{40}{100}$. Given that she has been assigned to the big group, the probability B is assigned to the big group is $\frac{39}{99}$. And so on. The probability they are all assigned to the big group is $$\frac{40}{100}\cdot\frac{39}{99}\cdot\frac{38}{98}\cdot\frac{37}{97}.$$

Write down similar expressions for the other three groups, and add. The denominators are all the same, so the computation is quick.

Remark: Your second lengthy expression gives the correct answer, and in fact fairly quickly simplifies to the answer one gets by using the procedure described above.

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