If a cricket team of 11 players is to be selected from 8 batsman, 6 bowlers, 4 all rounder and 2 wicket keepers, then
The numbr of selections when atmost 1 all rounder and 1 wicket keeper will play is

Options

A. `.^(4)C_(1).^(14)C_(10)+.^(2)C_(1).^(14)C_(10)+.^(4)C_(1) .^(2)C_(1).^(14)C_(9)+.^(14)C_(11)`
B. `.^(4)C_(1).^(14)C_(11)+.^(15)C_(11)`
C. `.^(4)C_(1).^(15)C_(10)+.^(15)C_(11)`
D. None of these

Answer

To solve the problem of selecting a cricket team of 11 players from a pool of 8 batsmen, 6 bowlers, 4 all-rounders, and 2 wicket-keepers, with the conditions that at most 1 all-rounder and at most 1 wicket-keeper can be included, we can break the problem down into cases based on the selections of all-rounders and wicket-keepers.

### Step-by-Step Solution

**Step 1: Identify the cases based on selections of all-rounders and wicket-keepers.**
- Case 1: 1 all-rounder and 1 wicket-keeper
- Case 2: 1 all-rounder and 0 wicket-keepers
- Case 3: 0 all-rounders and 1 wicket-keeper
- Case 4: 0 all-rounders and 0 wicket-keepers

**Step 2: Calculate the number of ways for each case.**

**Case 1: 1 all-rounder and 1 wicket-keeper**
- Choose 1 all-rounder from 4: \( \binom{4}{1} \)
- Choose 1 wicket-keeper from 2: \( \binom{2}{1} \)
- Choose the remaining 9 players from the remaining 14 players (8 batsmen + 6 bowlers): \( \binom{14}{9} \)

Total for Case 1:
\[ \text{Total}_1 = \binom{4}{1} \times \binom{2}{1} \times \binom{14}{9} \]

**Case 2: 1 all-rounder and 0 wicket-keepers**
- Choose 1 all-rounder from 4: \( \binom{4}{1} \)
- Choose 10 players from the remaining 14 players: \( \binom{14}{10} \)

Total for Case 2:
\[ \text{Total}_2 = \binom{4}{1} \times \binom{14}{10} \]

**Case 3: 0 all-rounders and 1 wicket-keeper**
- Choose 1 wicket-keeper from 2: \( \binom{2}{1} \)
- Choose 10 players from the remaining 14 players: \( \binom{14}{10} \)

Total for Case 3:
\[ \text{Total}_3 = \binom{2}{1} \times \binom{14}{10} \]

**Case 4: 0 all-rounders and 0 wicket-keepers**
- Choose 11 players from the remaining 14 players: \( \binom{14}{11} \)

Total for Case 4:
\[ \text{Total}_4 = \binom{14}{11} \]

**Step 3: Calculate the total number of selections.**
Now, we sum the totals from all four cases:
\[ \text{Total selections} = \text{Total}_1 + \text{Total}_2 + \text{Total}_3 + \text{Total}_4 \]
Substituting the values:
\[ \text{Total selections} = \left( \binom{4}{1} \times \binom{2}{1} \times \binom{14}{9} \right) + \left( \binom{4}{1} \times \binom{14}{10} \right) + \left( \binom{2}{1} \times \binom{14}{10} \right) + \left( \binom{14}{11} \right) \]

### Final Calculation
Now, we can compute the values:
- \( \binom{4}{1} = 4 \)
- \( \binom{2}{1} = 2 \)
- \( \binom{14}{9} = \binom{14}{5} = 2002 \)
- \( \binom{14}{10} = \binom{14}{4} = 1001 \)
- \( \binom{14}{11} = \binom{14}{3} = 364 \)

Now substituting these values:
\[ \text{Total selections} = (4 \times 2 \times 2002) + (4 \times 1001) + (2 \times 1001) + 364 \]
Calculating each term:
- \( 4 \times 2 \times 2002 = 16016 \)
- \( 4 \times 1001 = 4004 \)
- \( 2 \times 1001 = 2002 \)
- \( 364 = 364 \)

Now summing these:
\[ \text{Total selections} = 16016 + 4004 + 2002 + 364 = 20486 \]

Thus, the total number of selections when at most 1 all-rounder and 1 wicket-keeper will play is **20486**.