Question

The total number of ways of formation of five letter words from the letters of the word
I N D E P E N D E N T, is?
$
  {\text{A}}{\text{. 4200}} \\
  {\text{B}}{\text{. 3320}} \\
  {\text{C}}{\text{. 3840}} \\
  {\text{D}}{\text{. None of these}} \\
$

Answer 1

Hint – To find the total number of ways to form a 5 letter word, we calculate the number of letters in the given word. Look at the number of distinct letters and repeated letters from the given word and calculate the total number of possibilities of forming a 5 letter word using permutations and combinations.

Complete step-by-step answer:
Given data,
The word – I N D E P E N D E N T
There are a total of 11 letters in the word I N D E P E N D E N T.
Out of which there are 6 distinct letters, (I, N, D, E, P, and T).
And a total of (3 times N, 3 times E, 2 times D, I, P, T).
Now the number of different possibilities of forming a 5 letter word from these letters are –
All the letters are different,
The number of ways to pick 5 letters from a set of 6 letters is${}^{\text{6}}{{\text{C}}_{\text{5}}}$.
To form a 5 letter word, we have 5 spaces to be filled by 5 letters = 5! ways.
The number of ways of forming a word is =${}^{\text{6}}{{\text{C}}_{\text{5}}}$× 5! = 720.

2 letters are same, 3 letters are different
To have two letter same we have to pick for N, E or D =${}^3{{\text{C}}_1}$
After picking there are 5 letters from which 3 different letters are to be picked =${}^5{{\text{C}}_3}$
To form a 5 letter word, we have 5 spaces to be filled by 5 letters out of which two spaces are repeated =$\dfrac{{5!}}{{2!}}$ways.
The number of ways of forming a word =${}^3{{\text{C}}_1}$×${}^5{{\text{C}}_3}$×$\dfrac{{5!}}{{2!}}$= 1800 ways.

3 letters are same, 2 letters are different
To have three letter same we have to pick for N or E =${}^2{{\text{C}}_1}$
After picking there are 5 letters from which 2 different letters are to be picked =${}^5{{\text{C}}_2}$
To form a 5 letter word, we have 5 spaces to be filled by 5 letters out of which three spaces are repeated =$\dfrac{{5!}}{{3!}}$ways.
The number of ways of forming a word =${}^2{{\text{C}}_1}$×${}^5{{\text{C}}_2}$×$\dfrac{{5!}}{{3!}}$= 400 ways.

2 letters are one type of similar, 2 letters are other types of similar and 1 letter is different.
To have two, two letter repetitions we have to pick two letters between N, E or D =${}^3{{\text{C}}_2}$
After picking, there are 4 letters from which 1 letter is to be picked =${}^4{{\text{C}}_1}$
To form a 5 letter word, we have 5 spaces to be filled by 5 letters out of which there are two different two letter spaces repeated =$\dfrac{{5!}}{{2!2!}}$ways.
Total number of ways of forming a word =${}^3{{\text{C}}_2}$×${}^4{{\text{C}}_1}$×$\dfrac{{5!}}{{2!2!}}$= 360 ways.

3 letters are same, 2 letters are same
To have three letters same we have to pick between N and E =${}^2{{\text{C}}_1}$ways
After picking, we have two letters (N/E and D) out of which one has to be picked =${}^2{{\text{C}}_1}$ways
To form a 5 letter word, we have to fill 5 spaces out of which 3 are alike and 2 are alike =$\dfrac{{5!}}{{3!2!}}$ways.
Total number of ways of forming a word =${}^2{{\text{C}}_1}$×${}^2{{\text{C}}_1}$×$\dfrac{{5!}}{{3!2!}}$= 40 ways.

Hence the total number of ways of forming a 5 letter word from the word I N D E P E N D E N T is
= 720 + 1800 + 400 + 360 + 40 = 3320 ways.
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 5 letters and arranging them in each case and add all the possibilities for the answer.
If the order of arrangement matters then we call it a permutation, it is given by${}^{\text{n}}{{\text{P}}_{\text{r}}} = \dfrac{{{\text{n!}}}}{{\left( {{\text{n - r}}} \right)!}}$.
If the order of arrangement does not matter then we call it a combination, it is given by${}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n!}}}}{{\left( {{\text{n - r}}} \right)!{\text{r!}}}}$.
n! of a number = n (n-1) (n-2) …….. (n - (n-1)).
Example: 3! = 3×2×1.