Probability of neither person drawing a pair of gloves A question from my probability textbook:

A box contains $10$ pairs of gloves. A draws out a single glove, then B draws one, then A draws a second, then B draws a second. Show that the chance of neither drawing a pair is ${{290}\over{323}}$.

Here's what I did. The possible sequences that work here are RLRL, LLLL, RRRR, LRLR, and the probability of getting them is$${{2(10)(10)(9)(8) + 2(10)(9)(8)(7)}\over{20(19)(18)(17)}} = {{73}\over{323}},$$which does not equal the probability given in the book. So what did I do wrong? Is the book correct? Is there an alternative meaning of "neither drawing a pair" (that's the exact phrasing from the book) that I'm not aware of?
 A: I think the question means that each pair of gloves is different. So you draw a pair when you draw $R$ and $L$ of the same pair and not any $R$ and $L$.
First $A$ draws any glove. Then $B$ draws a single glove -
i) $B$ draws from the same pair that $A$ drew.
Probability is $\frac{1}{19}$ and that ensures none of them draw a pair.
ii) $B$ does not draw from the same pair.
Probability is $\frac{18}{19}$. Probability that now $A$ draws -
a) From the same pair as $B$ is $\frac{1}{18}$ and that ensures neither of them draw a pair.
b) Draws neither of own pair or of $B$ is $\frac{16}{18}$. Lastly $B$ does not draw a pair with probability $\frac{16}{17}$.
Hence the desired probability is,
$\frac{1}{19} + \frac{18}{19} \left[\frac{1}{18} + \frac{16}{18} \cdot \frac{16}{17}\right] = \frac{290}{323}$
A: Inclusion/exclusion seems to work well here.  I am assuming the pairs are different.
Let's say the event that $A$ draws a pair is $pair_A$ and the event that $B$ draws a pair is $pair_B$.
Then
$$P(pair_A) = P(pair_B) = \frac{10}{\binom{20}{2}} = \frac{1}{19}$$
and
$$P(pair_A \cap pair_B) = \frac{\binom{10}{2}\cdot 2!}{\binom{20}{2} \binom{18}{2}} = \frac{1}{323}$$
By inclusion/exclusion, the probability that neither $A$ nor $B$ draws a pair is
$$1 - P(pair_A) - P(pair_B) + P(pair_A \cap pair_B) = \frac{290}{323}$$
A: Another way to look at it.
We can as well let each draw her full quota in one swoop.
Suppose the first draws $\boxed{A|B}$ in $\binom{10}2\cdot2^2 = 180 $ ways
The next can draw $\boxed{A|B},\, \left[\boxed{A|X}\, \boxed{X|B}\right] \,or\; \boxed{X|Y}$
in $1 + \binom81 2^2 +\binom82 2^2= 145$ ways
Thus $Pr = \dfrac{180*145}{\binom{20}2\binom{18}2} = \dfrac{290}{323}$
