How many permutations of the letters of the w | vedclass

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(B) The word $ENDEANOEL$ has $9$ letters: $E, E, E, N, D, A, O, N, L$. The letters are $E(3), N(2), D(1), A(1), O(1), L(1)$.Total letters = $9$. We ne

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$2 	imes 5!$

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(B) The word $ENDEANOEL$ has $9$ letters: $E, E, E, N, D, A, O, N, L$. The letters are $E(3), N(2), D(1), A(1), O(1), L(1)$.Total letters = $9$. We need to arrange them such that the last $5$ positions do not contain $D, L,$ or $N$.This means the last $5$ positions must be filled by the remaining letters $E, E, E, A, O$.The number of ways to arrange $E, E, E, A, O$ in the last $5$ positions is $\frac{5!}{3!} = \frac{120}{6} = 20$.The remaining $4$ positions must be filled by the remaining letters $D, L, N, N$. The number of ways to arrange these is $\frac{4!}{2!} = \frac{24}{2} = 12$.Total permutations = $20 	imes 12 = 240$.Since $2 	imes 5! = 2 	imes 120 = 240$,the correct option is $B$.

How many permutations of the letters of the word $ENDEANOEL$ are there such that the last five positions do not contain $D, L,$ or $N$?

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