Question

The number of words that can be written using all the letters of the word “IRRATIONAL” is
a. \[\dfrac{{10!}}{{{{(2!)}^3}}}\]
b. \[\dfrac{{10!}}{{{{(2!)}^2}}}\]
c. \[\dfrac{{10!}}{{(2!)}}\]
d. \[10!\]

Answer 1

Hint: This is a general permutation and combination problem where we are using general permutation and combination theorems. Choosing the number of letters in the given word we will find the number of ways of distributing it. Then analyzing the options with respect to our answer will give us our desired result.

Complete step-by-step answer:
There are 10 letters in the word IRRATIONAL.
The letters are, I (2 times), R (2 times), A(2 times), T, O, N, L
Now, 10 letters can be distributed among themselves in the number of ways, \[10!\]
Now, it would be the answer if there is no repetition of letters.
So, we can see there are three letters who are repeating themselves twice,
So, for the letter I, we get, the number of ways it can distribute itself is, \[2!\],
And now, for the letter R, we get, the number of ways it can distribute itself is, \[2!\],
And again, for the letter A, we get, the number of ways it can distribute itself is, \[2!\],
So, now, the number of words that can be written using all the letters of “IRRATIONAL” is,
\[ = \dfrac{{10!}}{{2! \times 2! \times 2!}}\]
\[ = \dfrac{{10!}}{{{{(2!)}^3}}}\]
Hence, option (a) is correct.

Note: In this type of problem we need to find the letters which are repeating in the given word. As the letters are repeating twice we are dividing the total number by \[2!\], if it would have been 3, then we had to divide it by \[3!\]and find the result.
As the repeating letters are identical so their position doesn’t matter, hence we remove that case by dividing it by \[r!\].