Question

The number of arrangements of these letters of the word ‘CALCUTTA’.
${\text{A}}{\text{. 2520}} \\\ {\text{B}}{\text{. 5040}} \\\ {\text{C}}{\text{. 10080}} \\\ {\text{D}}{\text{. 40320}} \\\$

Answer 1

Hint – In order to find the number of arrangements of the letters from the given word, we consider the number of possibilities of picking the letters from the given word. The resulting word also has the same number of letters as the given word and that means there can be repetitions.

Complete step-by-step answer:
Given Data,
CALCUTTA
The total number of letters in the given word CALCUTTA is 8.
It has –
2 C’s,
2 T’s,
2 A’s,
U and L.
We have to find the number of ways we can arrange these 8 letters (including the repeated letters).
So the resulting word has 8 spaces to fill - _ _ _ _ _ _ _ _.
The number of ways we can arrange ‘n' objects in n different ways is given by ‘n!’. Here we have 8 letters to fill in 8 spaces, so the number of ways to arrange this is 8!
But in our case we have, 3 letters out of the 8 letters that are repeated twice,
We pick each of the repeated letters from its set of two repeated letters in${}^2{{\text{C}}_1}$ways.
So the number of ways of arranging is given by
$\Rightarrow \dfrac{{8!}}{{{}^2{{\text{C}}_1} \times {}^2{{\text{C}}_1} \times {}^2{{\text{C}}_1}}}$
$\Rightarrow \dfrac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 2 \times 2}} = 5040$
Hence Option B is the correct answer.

Note – In order to solve this type of question the key is to carefully list out all the possible ways we can arrange a word out of all possible ways we can pick from the given word.
We have to individually calculate the possibilities of picking 8 letters and arranging them by considering the repetitions for the answer.
If the order of arrangement does not matter then we call it a combination, it is given by${}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n!}}}}{{\left( {{\text{n - r}}} \right)!{\text{r!}}}}$.
Our case of selecting from the repeated numbers is a combination because we get the same output even if either of the repeated letters is picked first.
n! of a number is given by n (n-1) (n-2) …….. (n - (n-1)).