1. Classify the players

There are:

• $6$ bowlers
• $7$ batsmen
• $2$ wicketkeepers

We need to form a team of $11$ such that it contains:

• at least $4$ bowlers
• at least $5$ batsmen
• at least $1$ wicketkeeper

Let the selected numbers be:

• bowlers $= b$
• batsmen $= m$
• wicketkeepers $= w$

Then $b+m+w=11$ with constraints $b\ge 4,\quad m\ge 5,\quad w\ge 1$ Also, $b\le 6,\quad m\le 7,\quad w\le 2$

2. Find all possible compositions $(b,m,w)$

Since minimum required players are: $4+5+1=10$ we need $1$ extra player beyond the minimum.

So distribute this $1$ extra among bowlers, batsmen, wicketkeepers, respecting upper limits.

Possible cases:

• $(4,5,2)$
• $(4,6,1)$
• $(5,5,1)$

These are the only valid possibilities.

3. Count selections for each case

Case 1: $(b,m,w)=(4,5,2)$

Choose:

• $4$ bowlers from $6$: $\binom{6}{4}$
• $5$ batsmen from $7$: $\binom{7}{5}$
• $2$ wicketkeepers from $2$: $\binom{2}{2}$

Number of ways: $\binom{6}{4}\binom{7}{5}\binom{2}{2}=15\cdot 21\cdot 1=315$

Case 2: $(b,m,w)=(4,6,1)$

Choose:

• $4$ bowlers from $6$: $\binom{6}{4}$
• $6$ batsmen from $7$: $\binom{7}{6}$
• $1$ wicketkeeper from $2$: $\binom{2}{1}$

Number of ways: $\binom{6}{4}\binom{7}{6}\binom{2}{1}=15\cdot 7\cdot 2=210$

Case 3: $(b,m,w)=(5,5,1)$

Choose:

• $5$ bowlers from $6$: $\binom{6}{5}$
• $5$ batsmen from $7$: $\binom{7}{5}$
• $1$ wicketkeeper from $2$: $\binom{2}{1}$

Number of ways: $\binom{6}{5}\binom{7}{5}\binom{2}{1}=6\cdot 21\cdot 2=252$

4. Add all cases

Total number of valid teams: $315+210+252=777$

5. Final Answer

The number of ways is $\boxed{777}$

6. Comparison with stored correct answer

Stored correct answer = $777$

Our derived answer also equals $777$, so they agree.