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Problem: We have 40 books on a shelf randomly arranged. Three books are a series and have numbers - 1,2,3. Need to find the probability that they'll be arranged at ascending order, like 1 comes earlier than 2, and 2 earlier than 3, but they need not to be right after each other. Like this case is possible (1.....2......3)

Now I managed to find the probability when 123 come right after each other. It is about $0.00064$. But I can't find a strategy to solve the above mentioned one. Help would be very appreciated.

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    $\begingroup$ Please note that --ignoring the spaces in between them-- there are only six ways to order these three books: (123), (132), (213), (231), (312) and (321). Since the books are randomly arranged, each of this six orderings must have the same probability. Which is therefore $1/6$. So that is your answer. $\endgroup$ – M. Wind Oct 25 '18 at 4:37
  • $\begingroup$ @M.Wind You should post your comment as an answer. $\endgroup$ – N. F. Taussig Oct 25 '18 at 8:49
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Note that the symmetry argument of M. Wind produces an elegant solution.

Here, I present a solution "on combinatorical foot":

  • First, arrange the $37$ other books: $\color{blue}{37!}$
  • These $37$ books offer $38$ "slots" where to put the $3$ books in the given specific order: $\color{green}{38}$
  • In each "slot" $s_i$ $(i=1,\ldots, \color{green}{38})$ you can put $s_i =0,1,2,3$ books, such that $s_1 + \cdots +s_{\color{green}{38}} = \color{orange}{3}$. As the order of the books is fixed, you get as the number of ways (using the formula for combinations with repetitions): $\binom{\color{green}{n}+\color{orange}{k}-1}{\color{orange}{k}}=\color{blue}{\binom{40}{3}}$

It follows: $$P(\text{books of series appear in ascending order}) = \frac{\color{blue}{37!\cdot \binom{40}{3}}}{40!} = \frac{1}{6}$$

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  • $\begingroup$ You made a computational mistake. Note that your $n = 38$ and your $k = 3$, so $\binom{n + k - 1}{k} = \binom{38 + 3 - 1}{3} = \binom{40}{3}$. The answer reduces to $1/6$, which can be more easily obtained by using a symmetry argument. $\endgroup$ – N. F. Taussig Oct 25 '18 at 8:48
  • $\begingroup$ @N.F.Taussig: OOOps. Thanks. Sometimes adding and subtracting isn't that easy ;-) $\endgroup$ – trancelocation Oct 25 '18 at 8:56
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Please note that --ignoring the spaces in between them-- there are only six ways to order these three books: (123), (132), (213), (231), (312) and (321). Since the books are randomly arranged, each of this six orderings must have the same probability. Which is therefore 1/6. So that is your answer.

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