A team of 11 players with at least 3 bowlers and 1 wicket keeper is to be formed from 16 players; 4 are bowlers; 2 are wicket keepers. 
Out of 16 players in a cricket team, 4 are bowlers and 2 are wicket keepers. A team of 11 players with at least 3 bowlers and 1 wicket keeper is to be formed. Find the number of ways the team can be selected.        

My solution:
Choosing 3 bowlers out of 4: $\binom43$
Choosing 1 wicket keeper out of 2: $\binom21$
Choosing the remaining 4 players out of 12: $\binom{12}{7}$   
Hence, total permutations=$\binom43\binom21\binom{12}{7}=6336$   
Given solution:
Number of bowlers=4
Number of wicket keepers=2
Total required selection=$\binom43\binom21\binom{10}{7} +\binom44\binom21\binom{10}{6}+\binom43\binom22\binom{10}{6}+\binom44\binom22\binom{10}{5}=2472$     
I feel my solution is just a simpler version of what the book gives. But why is then my answer coming different and where have I gone wrong in my approach?

Firstly, it has to be solved through combination.
You have used Permutation.
Second, the n must add up to 16 , but yours adds up to 18.
Third, the problem specifies "at least 3 bowlers and atleast 1 wk", which means you have to try out combinations of 
( n=4 combination with r= 3 and r=4 )for bowlers
(n=2 combination with r= 1 and r=2) for wicket keepers
Then the remaining players have to be accounted for in each combination.
So you'll get 
(4C3 * 2C1*10C7) + (4C3* 2C2*10C6) + (4C4*2C1*10C6) + (4C4* 2C2*10C5)
=960 + 840 + 252 +420
=2472
