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(A) The word "$IRRATIONAL$" contains $10$ letters in total.Counting the frequency of each letter:$I$ appears $2$ times.$R$ appears $2$ times.$A$ appea

Vedclass · Accepted Answer

$\frac{10 !}{(2 !)^{3}}$

Vedclass · Answer

(A) The word "$IRRATIONAL$" contains $10$ letters in total.Counting the frequency of each letter:$I$ appears $2$ times.$R$ appears $2$ times.$A$ appears $2$ times.$T, O, N, L$ appear $1$ time each.Using the formula for permutations of a multiset, the total number of arrangements is given by $\frac{n!}{n_1! n_2! n_3! \dots n_k!}$.Here, $n = 10$, $n_1 = 2$ (for $I$), $n_2 = 2$ (for $R$), and $n_3 = 2$ (for $A$).Therefore, the total number of words $= \frac{10!}{2! 2! 2!} = \frac{10!}{(2!)^3}$.

The number of words that can be written using all the letters of the word "$IRRATIONAL$" is

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