In how many ways can four persons be seated in $9$ chairs?

4!

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9!

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^{9} P_{4}

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^{9} C_{4}

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Solution

The correct option is B $^{9} P_{4}$
Out of these 9 chairs select $4$ out of them .
These $4$ are then permuted
$=^{9} C_{4} \times 4!$
$=^{9} P_{4}$

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