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Permutation under Restriction

Number of per...

Question

Number of permutations that can be formed with the letters of the word "TRIANGLE" is

A

8!

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B

\frac{8!}{2!}

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C

\frac{8!}{3!}

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D

\frac{8!}{{(2!)}^{2}}

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Solution

The correct option is A 8!
TRIANGLE
$\to$ This word has 8 distinct letters. So, the total no. of permutations is $= 8!$ .

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