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In how many ways can the letters of the word CORPORATION be arranged so that vowels always occupy even places?
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The correct answer is:
7200
CORPORATION is 11 letter word. It has 5 vowels $(\mathrm{O}, \mathrm{O}, \mathrm{O}, \mathrm{A}, \mathrm{I})$ and 6 consonants $(\mathrm{C}, \mathrm{R}$
$\mathrm{P}, \mathrm{R}, \mathrm{T}, \mathrm{N})$
In 11 letters, there are 5 even places $\left(2^{\text {nd }}, 4^{\text {th }}, 6^{\text {th }}, 8^{\text {th }}\right.$ and $10^{\text {th }}$ positions)
5 vowels can take 5 even places in $\frac{5 !}{3 !}$ ways
$(\because$ Since $\mathrm{O}$ is repeated thrice)
Similarly, 6 consonasts can take 6 odd places in $\frac{6 !}{2 !}$
ways. $(\because \mathrm{R}$ is repeated twice)
$\therefore \quad$ Total number of ways $=\frac{5 !}{3 !} \times \frac{6 !}{2 !}=20 \times 360=7200$
$\mathrm{P}, \mathrm{R}, \mathrm{T}, \mathrm{N})$
In 11 letters, there are 5 even places $\left(2^{\text {nd }}, 4^{\text {th }}, 6^{\text {th }}, 8^{\text {th }}\right.$ and $10^{\text {th }}$ positions)
5 vowels can take 5 even places in $\frac{5 !}{3 !}$ ways
$(\because$ Since $\mathrm{O}$ is repeated thrice)
Similarly, 6 consonasts can take 6 odd places in $\frac{6 !}{2 !}$
ways. $(\because \mathrm{R}$ is repeated twice)
$\therefore \quad$ Total number of ways $=\frac{5 !}{3 !} \times \frac{6 !}{2 !}=20 \times 360=7200$
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