TITLE: Probability of correct books ordering
QUESTION [3 upvotes]: Suppose I have 10 books in heap and I want to put 6 of them into 
bookshelf. Also each book has a number on the cover (from 1 to 10). How to calculate probability of situation where books with numbers 1,2,3,4 appears on the bookshelf and in the increasing order? (Other 2 of 6 chosen books could be in random order)
There is my attempt to solve it: 
Firstly, calculate the probability of the appearance of the right 4 books among the selected 6. This is $$p_1=\frac{6 \choose 2}{10 \choose 6} = \frac{1}{14}$$
Secondly, find probability of situation, when books 1-4 are in increasing order. $$p_2 = \frac{2 \cdot {6 \choose 2}}{6!} = \frac{1}{24}$$
Finally, my answer is $$p_1 \cdot p_2 = \frac{1}{336}$$ 
But in my textbook ansew is $$\frac{1}{36}$$. What is wrong with my solution? Or it is typo in the book?

REPLY [0 votes]: I obtain the same your result indeed we have

overall cases $$6!\binom{10}{6}$$
situations where books with numbers $1,2,3,4$ appears on the bookshelf
and in the increasing order

$$2! \binom{6}{4}\binom{6}{2}$$
therefore result should be
$$\frac{2!\binom{6}{4}\binom{6}{2}}{6!\binom{10}{6}}=\frac1{336}$$
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