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How many different sequences can be made from the numbers $ \{0, 1, 2\} $ where each number has exactly 20 occurrences?

I think it's the case of repeated permutations. If I interpret the problem correctly,

$$ 000111222 $$ $$ 012210102 $$

are some possible sequences where each number is represented 3 times.

Looking at the definition in the textbook the correct answer seems to be

$$ \frac{(3\cdot 20)!}{(20!)^3} $$

But I have no idea how be sure.

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  • $\begingroup$ You have three numbers 0,1 and 2 so if each of them has 20 occurrences that means you're going to have a total of 60 elements. $\endgroup$ – Module Jan 18 '14 at 22:14
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Your answer is correct!

Think of a $60$-digit line where you choose $20$ places for each number. This choice is done with combination because the arrangement of the identical numbers do not matter.

The answer thus is: $\Large\binom{60}{20}\cdot\binom{40}{20}\cdot\binom{20}{20}$

Which is same as $\Large\frac{60!}{40!\cdot20!}\cdot\frac{40!}{20!\cdot20!}\cdot\frac{20!}{20!\cdot0!}=\frac{60!}{20!\cdot20!\cdot20!}$

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  • $\begingroup$ So would $ \frac{(n\cdot m)!}{(m!)^n} $ a general solution for all similar problems with n numbers each with m occurrences? $\endgroup$ – kdani Jan 18 '14 at 22:22
  • $\begingroup$ Yes. If they occur unequally, for example the first one $n_1$ times, the second one $n_2$ times,... and the $r$th one $n_r$ times, the general solution becomes $\Large\frac{(n_1+n_2+\dots+n_r)!}{(n_1)!\cdot(n_2)!\cdots(n_r)!}$ $\endgroup$ – Zafer Cesur Jan 18 '14 at 22:27

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