Consider all possible permutations of the let | vedclass

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(A) The word $ENDEANOEL$ has $9$ letters: $E, E, E, N, N, D, A, O, L$. $(A)$ Treating $ENDEA$ as a single block,we have the block $(ENDEA)$ and the re

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$(A) \rightarrow (s); (B) \rightarrow (r); (C) \rightarrow (p); (D) \rightarrow (q)$

Vedclass · Answer

(A) The word $ENDEANOEL$ has $9$ letters: $E, E, E, N, N, D, A, O, L$. $(A)$ Treating $ENDEA$ as a single block,we have the block $(ENDEA)$ and the remaining letters $N, O, E, L$. Total items $= 5$. The number of permutations is $5! = 120$. This matches $(p)$. $(B)$ Total letters $= 9$. $E$ occurs $3$ times. If $E$ is at the first and last position,we have $7$ positions left to fill with $E, N, N, D, A, O, L$. The number of ways is $\frac{7!}{2!} = \frac{5040}{2} = 2520 = 21 	imes 120 = 21 	imes 5!$. This matches $(s)$. $(C)$ None of $D, L, N$ in the last $5$ positions means $D, L, N$ must be in the first $4$ positions. The letters are $E, E, E, N, N, D, A, O, L$. The first $4$ positions must contain $D, L, N$ and one $E$. Ways $= \frac{4!}{2!} = 12$. The last $5$ positions must contain the remaining $E, E, A, O$. Ways $= \frac{5!}{3!} = 20$. Total $= 12 	imes 20 = 240 = 2 	imes 120 = 2 	imes 5!$. This matches $(q)$. $(D)$ $A, E, O$ occur only in odd positions $(1, 3, 5, 7, 9)$. There are $5$ odd positions. We have $3$ $E$'s,$1$ $A$,$1$ $O$. Total $5$ letters. Ways to arrange these in $5$ odd positions $= \frac{5!}{3!} = 20$. The remaining $4$ letters $(N, N, D, L)$ in $4$ even positions $= \frac{4!}{2!} = 12$. Total $= 20 	imes 12 = 240 = 2 	imes 5!$. This matches $(q)$.

$Column I$	$Column II$
$(A)$ The number of permutations containing the word $ENDEA$ is	$(p)$ $5!$
$(B)$ The number of permutations in which the letter $E$ occurs in the first and the last positions is	$(q)$ $2 \times 5!$
$(C)$ The number of permutations in which none of the letters $D, L, N$ occurs in the last five positions is	$(r)$ $7 \times 5!$
$(D)$ The number of permutations in which the letters $A, E, O$ occur only in odd positions is	$(s)$ $21 \times 5!$

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