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I have 10 students in the class, and have 18 numbers.

Every student will get 9 random numbers from this set.

$$\begin{array}{c|c|c} 7&10&1\\\hline 9&2&17\\\hline 8&5&7 \end{array}$$

If I'm not wrong, there are $9! = 362880$ possibilities the students have the same order if they have the same number as another student.

How can I calculate the possibilities some students have the same number and same order as another student?

grg
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Anthony Dekoci
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1 Answers1

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Since the $9$ random numbers given to a student are not necessarily distinct, it follows that there are $N=18^9$ ways. The probability that at least two students get the same $9$ numbers in the same order is $$1-\frac{N(N-1)\cdots (N-9)}{N^{10}}$$ that is the complement of the probability that they are all distinct.

Robert Z
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