1. Interpret the condition carefully

We have:

• $7$ batsmen, among whom one specific batsman must be included as captain.
• $6$ bowlers, among whom one specific bowler must be included as vice-captain.

A team of $10$ players is to be selected such that:

• at least $4$ batsmen are included,
• at least $4$ bowlers are included,
• the specified captain and vice-captain must be included.

Since only batsmen and bowlers are available, if the team has $b$ batsmen and $w$ bowlers, then $b+w=10$ with $b\ge 4, \quad w\ge 4.$

2. Possible compositions of the team

From $b+w=10$ and both at least $4$, the possibilities are:

• $(4,6)$
• $(5,5)$
• $(6,4)$

We now count each case, ensuring the fixed captain batsman and fixed vice-captain bowler are included.

3. Case 1: 4 batsmen and 6 bowlers

• Since all $6$ bowlers are selected, and the vice-captain bowler is among them automatically, bowlers can be chosen in $\binom{5}{5}=1$ way after fixing the vice-captain.

• For batsmen, one fixed captain is already included. We need $3$ more from the remaining $6$ batsmen: $\binom{6}{3}=20.$

So total ways in this case: $20 \cdot 1=20.$

4. Case 2: 5 batsmen and 5 bowlers

• Batsmen: captain is fixed, so choose $4$ more from remaining $6$: $\binom{6}{4}=15.$

• Bowlers: vice-captain is fixed, so choose $4$ more from remaining $5$: $\binom{5}{4}=5.$

So total ways in this case: $15 \cdot 5=75.$

5. Case 3: 6 batsmen and 4 bowlers

• Batsmen: captain is fixed, so choose $5$ more from remaining $6$: $\binom{6}{5}=6.$

• Bowlers: vice-captain is fixed, so choose $3$ more from remaining $5$: $\binom{5}{3}=10.$

So total ways in this case: $6 \cdot 10=60.$

6. Add all cases

Total number of teams: $20+75+60=155.$

7. Compare with options

The correct option is $\boxed{155}$ which is Option C.

8. Comparison with stored answer

Stored correct answer: C

Our derived answer is also C.