From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsman and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is

We need to select 10 players from 7 batsmen and 6 bowlers, including a designated captain (batsman) and vice-captain (bowler), with at least 4 batsmen and 4 bowlers total.

Since the captain (1 batsman) and vice-captain (1 bowler) are already included, we need 8 more players from the remaining 6 batsmen and 5 bowlers, with at least 3 more batsmen and 3 more bowlers.

The possible distributions of (additional batsmen, additional bowlers) are:

Case 1: (3, 5): $$\binom{6}{3} \times \binom{5}{5} = 20 \times 1 = 20$$

Case 2: (4, 4): $$\binom{6}{4} \times \binom{5}{4} = 15 \times 5 = 75$$

Case 3: (5, 3): $$\binom{6}{5} \times \binom{5}{3} = 6 \times 10 = 60$$

Total number of selections = $$20 + 75 + 60 = 155$$.

Hence, the correct answer is Option B.