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$5$ people will choose randomly and independently an integer between $1$ and $200$. What is the probability that there will be exactly $3$ people who choose the same number?

Here is my solution:

$$\frac{{200 \choose 2}{2\choose 1}{5 \choose 3}+{200 \choose 3}{3\choose 1}{5 \choose 3}2}{200^5}$$

is it correct? Any suggestions?

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    $\begingroup$ How did you come up with this answer? Would you explain? $\endgroup$ – Vinyl_cape_jawa Feb 28 at 14:13
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    $\begingroup$ The answer is correct. $\endgroup$ – NCh Feb 28 at 14:26
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    $\begingroup$ The answer is correct. The first term describes the number of ways in which three people match with one number and the remaining two people match while the second term describes the analogous number of ways where instead the remaining two had different numbers. $\endgroup$ – JMoravitz Feb 28 at 14:27
  • $\begingroup$ The answer can be found more directly, as shown below, but it will be the same answer otherwise just appearing in a different form. $\endgroup$ – JMoravitz Feb 28 at 14:28
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Let's fix first the three people that will choose the same number. The first of them has $200$ possibilities, and the two others have no possibilities, they have to choose the same. Finally, the two last people can choose among $199$ numbers, they can't choose the same numbers as the three firsts. So there are $200 \times 199 \times 199$ possibilities.

Now, you have to integrate the fact that there are ${5 \choose 3}$ ways to choose the three people concerned. So the total number of cases is equal to ${5 \choose 3} \times 200 \times 199^2$.

Finally, you get that the probability is equal to $\frac{{5 \choose 3} \times 200 \times 199^2}{200^5} \simeq 0.02 \%$.

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