Find the number of arrangements of the letter | vedclass

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Question

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

Vedclass · Accepted Answer

$138600$

Vedclass · Answer

(A) The word $INDEPENDENCE$ contains $12$ letters: $I(1), N(3), D(2), E(4), P(1), C(1)$.To find the number of arrangements starting with $P$,we fix $P$ at the first position.Now,we need to arrange the remaining $11$ letters: $I(1), N(3), D(2), E(4), C(1)$.The number of arrangements is given by the formula $\frac{n!}{n_1! n_2! n_3! ...}$.Here,$n = 11$,$n_1 = 3$ (for $N$),$n_2 = 2$ (for $D$),and $n_3 = 4$ (for $E$).Number of arrangements $= \frac{11!}{3! 2! 4!} = \frac{39916800}{6 	imes 2 	imes 24} = \frac{39916800}{288} = 138600$.

Find the number of arrangements of the letters of the word $INDEPENDENCE$. In how many of these arrangements do the words start with $P$?

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