How many distinct words can be formed using t | vedclass

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(C) The word $ARRANGE$ contains $7$ letters: $A, R, R, A, N, G, E$. Here,$A$ appears $2$ times,$R$ appears $2$ times,and $N, G, E$ appear $1$ time eac

Vedclass · Accepted Answer

$900$

Vedclass · Answer

(C) The word $ARRANGE$ contains $7$ letters: $A, R, R, A, N, G, E$. Here,$A$ appears $2$ times,$R$ appears $2$ times,and $N, G, E$ appear $1$ time each. Total number of arrangements = $\frac{7!}{2! 	imes 2!} = \frac{5040}{4} = 1260$. To find the number of arrangements where two $A$'s do not occur together,we use the gap method. First,arrange the remaining letters $R, R, N, G, E$ in $\frac{5!}{2!} = \frac{120}{2} = 60$ ways. These $5$ letters create $6$ gaps (including ends): $\_ R \_ R \_ N \_ G \_ E \_$. We need to place $2$ $A$'s in these $6$ gaps. The number of ways to do this is $^6C_2 = \frac{6 	imes 5}{2} = 15$. Total arrangements where $A$'s are not together = $60 	imes 15 = 900$.

How many distinct words can be formed using the letters of the word $ARRANGE$ such that the two $A$'s do not occur together?

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