The number of arrangements of the letters of | vedclass

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(C) The word $ARRANGEMENT$ has $11$ letters: $A(2), R(2), N(2), E(2), G(1), M(1), T(1)$.Total arrangements $= \frac{11!}{2! \cdot 2! \cdot 2! \cdot 2!

Vedclass · Accepted Answer

$\frac{9}{16}(10!)$

Vedclass · Answer

(C) The word $ARRANGEMENT$ has $11$ letters: $A(2), R(2), N(2), E(2), G(1), M(1), T(1)$.Total arrangements $= \frac{11!}{2! \cdot 2! \cdot 2! \cdot 2!} = \frac{11!}{16}$.To find the arrangements where two $E$s do not occur together,we subtract the arrangements where they do occur together from the total.Treating the two $E$s as a single unit $(EE)$,we have $10$ items to arrange: $A(2), R(2), N(2), G(1), M(1), T(1), (EE)(1)$.Number of arrangements with $E$s together $= \frac{10!}{2! \cdot 2! \cdot 2!} = \frac{10!}{8}$.Number of arrangements with $E$s not together $= \frac{11!}{16} - \frac{10!}{8} = \frac{11 \cdot 10!}{16} - \frac{2 \cdot 10!}{16} = \frac{9 \cdot 10!}{16} = \frac{9}{16}(10!)$.

The number of arrangements of the letters of the word $ARRANGEMENT$ in which two $E$s do not occur adjacently is

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