A car will hold 2 persons in the front seat a | vedclass

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(B) The car has $3$ seats in total: $2$ in the front and $1$ in the rear. Since the driver must be in the front seat,we first select $1$ driver from t

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$20$

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(B) The car has $3$ seats in total: $2$ in the front and $1$ in the rear. Since the driver must be in the front seat,we first select $1$ driver from the $2$ eligible persons,which can be done in $^2C_1$ ways. After selecting the driver,we need to fill the remaining $2$ seats (one front,one rear) from the remaining $5$ persons. The number of ways to select $2$ persons from $5$ is $^5C_2$ ways. Since the order of the remaining $2$ passengers in the remaining $2$ seats matters (front vs rear),we multiply by $2!$ to account for their arrangement. Alternatively,we select the driver in $^2C_1$ ways,then select $1$ person for the remaining front seat from $5$ in $^5C_1$ ways,and finally $1$ person for the rear seat from the remaining $4$ in $^4C_1$ ways. Total ways = $^2C_1 	imes ^5C_1 	imes ^4C_1 = 2 	imes 5 	imes 4 = 40$. Wait,if the seats are distinct,the calculation is $2 	imes 5 	imes 4 = 40$. If the question implies the $2$ front seats are identical and the rear seat is distinct,the calculation is $^2C_1 	imes ^5C_2 = 2 	imes 10 = 20$.

$A$ car will hold $2$ persons in the front seat and $1$ person in the rear seat. If among $6$ persons $2$ can drive,then the number of ways in which the car can be filled is

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