At Least 4 Batsmen & 4 Bowlers? Count Selections FAST ⚡

 

❓ Question

From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include:

✔ at least 4 batsmen
✔ at least 4 bowlers
✔ and MUST include one batsman (captain) and one bowler (vice-captain)

Find the total number of ways such a team can be selected.

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🖼️ Question Image

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✍️ Short Solution

This is a classic combinations + constraints problem.

👉 Since captain (batsman) and vice-captain (bowler) must be selected, we include them first — then count valid ways to choose the remaining players abiding by the minimum batsman–bowler rule.

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🔹 Step 1 — Fix the compulsory players

We must include:

• 1 batsman (captain)

• 1 bowler (vice-captain)

So:

$$\text{Players already selected} = 2$$

Remaining to choose:

$$10 - 2 = 8$$

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🔹 Step 2 — Update available players

After fixing the captain & vice-captain:

• Remaining batsmen = $7 - 1 = 6$
  

• Remaining bowlers = $6 - 1 = 5$
  

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🔹 Step 3 — Apply “at least 4” condition

Total batsmen in final team ≥ 4

Since 1 batsman is fixed, we need:

$$\text{Extra batsmen } b \ge 3$$

Similarly for bowlers:

$$\text{Extra bowlers } w \ge 3$$

And total chosen now must be 8:

$$b + w = 8$$

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🔹 Step 4 — Solve possible distributions

List valid integer solutions satisfying all conditions:

Extra Batsmen $b$	Extra Bowlers $w$	Valid?
3	5	✔
4	4	✔
5	3	✔
6	2	❌ (bowlers < 4)

So only three valid cases.

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🔹 Step 5 — Count selections case-wise

We use combinations:

$${}^nC_r = \binom{n}{r}$$

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✅ Case 1: $b = 3,\ w = 5$

$$\binom{6}{3} \times \binom{5}{5} = 20 \times 1 = 20$$

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✅ Case 2: $b = 4,\ w = 4$

$$\binom{6}{4} \times \binom{5}{4} = 15 \times 5 = 75$$

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✅ Case 3: $b = 5,\ w = 3$

$$\binom{6}{5} \times \binom{5}{3} = 6 \times 10 = 60$$

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🔹 Step 6 — Add all valid selections

$$20 + 75 + 60 = 155$$

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✅ Final Answer

$$\boxed{155}$$