A team has nine players. How many possible batting orders are there?
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The 1st player can be placed in any one of the 9 positions.
The 2nd player can be placed in any one of the 8 positions (since the 1st player has already occupied 1 position.
Likewise the 9th player can be placed in the remaining 1 positon.
So no. Of batting order =9×8×7×6×5×4×3×2×1=362880
The 2nd player can be placed in any one of the 8 positions (since the 1st player has already occupied 1 position.
Likewise the 9th player can be placed in the remaining 1 positon.
So no. Of batting order =9×8×7×6×5×4×3×2×1=362880
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