Six 'X's have to be placed in the squares of | vedclass

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(C) The figure consists of $8$ squares arranged in three rows: top row ($2$ squares),middle row ($4$ squares),and bottom row ($2$ squares).We need to

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$26$

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(C) The figure consists of $8$ squares arranged in three rows: top row ($2$ squares),middle row ($4$ squares),and bottom row ($2$ squares).We need to place $6$ '$X$'s in these $8$ squares.The total number of ways to place $6$ '$X$'s in $8$ squares is $^8C_6 = \frac{8 	imes 7}{2 	imes 1} = 28$.We must ensure each row contains at least one '$X$'.Let $R_1, R_2, R_3$ be the top,middle,and bottom rows respectively.Total squares = $2 + 4 + 2 = 8$.If the top row $(R_1)$ has no '$X$',then all $6$ '$X$'s must be placed in the remaining $8 - 2 = 6$ squares. This can be done in $^6C_6 = 1$ way.If the bottom row $(R_3)$ has no '$X$',then all $6$ '$X$'s must be placed in the remaining $8 - 2 = 6$ squares. This can be done in $^6C_6 = 1$ way.Note that it is impossible for both the top and bottom rows to have no '$X$' simultaneously,as that would require placing $6$ '$X$'s in only $4$ squares,which is impossible.Thus,the number of invalid ways is $1 + 1 = 2$.The required number of ways is $28 - 2 = 26$.

Six '$X$'s have to be placed in the squares of the figure such that each row contains at least one '$X$'. In how many different ways can this be done?

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