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(C) The word $ATRAPATRAM$ contains $10$ letters: $A(4), T(2), R(2), P(1), M(1)$.$(i)$ If all $A$'s are together,we treat the block $(AAAA)$ as one uni

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$\frac{6!}{2!2!} 	imes 42$

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(C) The word $ATRAPATRAM$ contains $10$ letters: $A(4), T(2), R(2), P(1), M(1)$.$(i)$ If all $A$'s are together,we treat the block $(AAAA)$ as one unit. We then have $6$ other letters plus this $1$ unit,totaling $7$ items. The number of arrangements is $x = \frac{7!}{2!2!}$.(ii) If no two $A$'s are together,we first arrange the remaining $6$ letters $(T, T, R, R, P, M)$,which can be done in $\frac{6!}{2!2!}$ ways. There are $7$ gaps created by these $6$ letters. We choose $4$ gaps for the $4$ $A$'s in $^7C_4$ ways. Thus,$y = ^7C_4 	imes \frac{6!}{2!2!}$.Therefore,$x+y = \frac{7!}{2!2!} + ^7C_4 	imes \frac{6!}{2!2!}$.$x+y = \frac{7 	imes 6!}{2!2!} + 35 	imes \frac{6!}{2!2!} = \frac{6!}{2!2!} (7 + 35) = \frac{6!}{2!2!} 	imes 42$.

If $x$ and $y$ represent the number of arrangements of the letters of the word $ATRAPATRAM$ such that $(i)$ all $A$'s are together and $(ii)$ no two $A$'s are together respectively,then $x+y$ is equal to:

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