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(D) The word $LINEAR$ has $6$ distinct letters: $L, I, N, E, A, R$. We want to arrange these letters such that $E$ and $A$ are always together,but $N$

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$144$

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(D) The word $LINEAR$ has $6$ distinct letters: $L, I, N, E, A, R$. We want to arrange these letters such that $E$ and $A$ are always together,but $N$ and $R$ are not together. First,treat $(EA)$ as a single unit. Now we have $5$ units: $L, I, N, R, (EA)$. The number of ways to arrange these $5$ units is $5! = 120$. Since $E$ and $A$ can be arranged within their unit in $2! = 2$ ways,the total number of arrangements where $E$ and $A$ are together is $120 	imes 2 = 240$. Next,we find the number of arrangements where $E$ and $A$ are together $AND$ $N$ and $R$ are also together. Treat $(EA)$ as one unit and $(NR)$ as another unit. Now we have $4$ units: $L, I, (EA), (NR)$. The number of ways to arrange these $4$ units is $4! = 24$. Within their units,$E$ and $A$ can be arranged in $2! = 2$ ways,and $N$ and $R$ can be arranged in $2! = 2$ ways. So,the number of arrangements where both $(EA)$ and $(NR)$ are together is $24 	imes 2 	imes 2 = 96$. Finally,the number of ways where $E$ and $A$ are together but $N$ and $R$ are $NOT$ together is $240 - 96 = 144$.

The number of ways of arranging the letters of the word $LINEAR$ so that the letters $N$ and $R$ do not come together and $E$ and $A$ come together is

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