Arrangements of cricked players (statistics question)

A cricket team of 11 players is to be chosen from 21 players consisting of 10 batsmen, 9 bowlers, 2 wicketkeepers. The team must include 5 batsmen, 4 bowlers and 1 wicketkeepers.

Find the number of ways in which the team can be chosen.

Please give detailed calculations and comment on why and what you are doing.

you must be careful to avoid overcounting !
in the order batsman-bowler-wicketkeeper, permissible combos are
6-4-1, 5-5-1, and 5-4-2

# of ways
= 10c6*9c4*2c1 + 10c5*9c5*2c1 + 10c5*9c4*2c2 = 148,176 <----

So you need to choose 5 batsmen from the 10 available, 4 bowlers from the 9 available, 1 wicketkeeper from the 2 available and (11 - 5 - 4 - 1) = 1 more player from the (21 - 10) = 11 remaining. This can be done in (10C5)*(9C4)*(2C1)*(11C1) = 252*126*2*11 = 698544 ways.

Note that (nCr) = n!/((n-r)!*r!) is the formula for the number of combinations of r objects from a collection of n objects.

Edit: M3 is right; my approach resulted in over-counting. :)

Something's wrong with this problem, since 10 + 9 + 2 = 21 but 5 + 4 + 1 is not 11. This means that in the last step we have to choose 11 - 5 - 4 - 1 = 1 remaining player from the 21 - 10 - 9 - 2 = 0 available players who are neither batsmen nor bowlers nor wicketkeepers. This is not possible!