Find the number of arrangements of the letter | vedclass

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(A) The word $INDEPENDENCE$ has $12$ letters: $I(2), N(3), D(2), E(4), P(1)$.To find the number of arrangements starting with $I$ and ending with $P$,

Vedclass · Accepted Answer

$12600$

Vedclass · Answer

(A) The word $INDEPENDENCE$ has $12$ letters: $I(2), N(3), D(2), E(4), P(1)$.To find the number of arrangements starting with $I$ and ending with $P$,we fix $I$ at the first position and $P$ at the last position.Remaining letters are $10$: $N(3), D(2), E(4), I(1), N(0)$ (Wait,let us re-count: $I$ is used once,$P$ is used once).Remaining letters: $I(1), N(3), D(2), E(4)$.Total remaining letters = $1 + 3 + 2 + 4 = 10$.The number of arrangements is given by the formula $\frac{n!}{n_1! n_2! n_3! n_4!}$.Number of arrangements $= \frac{10!}{1! 3! 2! 4!} = \frac{3628800}{1 	imes 6 	imes 2 	imes 24} = \frac{3628800}{288} = 12600$.

Find the number of arrangements of the letters of the word $INDEPENDENCE$. In how many of these arrangements do the words begin with $I$ and end in $P$?

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