In a bridge game, North and South have 9 spades between them. Find the probability that either East or West has no spades. I am a complete novice in playing card games and I am unable to solve problems of this type . However I have tried to solve it as shown below:

Firstly we select $9$ cards out of $13$ spades and $17$ non spade cards out of $39$ non spade cards. These make a total of $26$ cards. Now we try to divide these $26$ cards between North and South. Now for each of these possible cases we try to choose $13$ non spade cards from $22$ non spade cards for either of East or West . Two such ways shall be possible i.e. either for East or for West. So the total number of favourable outcomes(m):

13C9 ×39C17×((26!)/(13!)^2)×22C13×2!

Total number of possible outcomes(n):
(52!)/(13!)^4

Required probability $=\frac{m}{n}\approx7.036337939\times10^{-3}$

But the answer given in book is not matching.( It is given as $=\frac{11}{115}$ , I am not sure whether the answer given is book is wrong or that of mine is wrong.)

Please help me out.
We have 4 spades and 22 others to distribute between East and West.  Let's suppose we deal the cards out, one by one, with the first 13 going to East and the second 13 going to West.  We will consider the order of the cards to be significant, so this can be done in $26!$ ways, all of which are equally likely.  

We want to count the number of sequences in which East or West gets all 4 spades.  There are $2$ ways to choose East/West, then there are $\binom{13}{4}$ ways to choose the locations of the spades within the chosen hand, $4!$ ways to sequence the spades, and $22!$ ways to sequence all the remaining cards.  So all together, there are 
$$2 \times \binom{13}{4} \times 4! \times 22!$$ 
ways to sequence the cards in which either East or West gets all the spades.

So the probability that either East or West gets all the spades is 
$$\frac{2 \times \binom{13}{4} \times 4! \times 22!}{26!} = \frac{11}{115}$$
Since this problem asks only about spades, there is no reason to say anything about "non-spade cards".  If North and South, together, have 9 spades then East and West, together, have 13- 9= 4 spades.  The probability any one of those 4 spades is in East's hand is 1/2.  The probability all 4 are in East's hand is $(1/2)^4= 1/16$.  Similarly the probability all 4 are in West's hand is $(1/2)^4= 1/16$.  The probability that "either East or West has no spades" which is the same as the probability "either East has all 4 remaining spades or West has all 4 remaining spades" is 1/16+ 1/16= 2/16= 1/8.