Question

How many words with or without meaning of three distinct letters of the English alphabet are there?

Answer 1

Hint: Here we will use the concept of finding all the possible outcomes which can be there.
We will arrange the letters in such a way that no letter gets repeated and only 3 letters are used at a time to form a word.

Complete step-by-step answer:
We will start by determining the number of letters in English alphabets.
There are 26 letters in English alphabets.
Now we need to form a three letter word out of 26 letters such that no letter is repeated.
The letters cannot be repeated because it is given in the question that the letters should be distinct.
So, we will start by selecting any three letters such as A, B, C out of the total number of letters i.e. 26 and form all possible words that can be made using these letters.
The possible words that can be made using these letters are:
ABC, ACB, BAC, BCA, CAB, CBA.
Now since it is impossible to make such pairs of letters and then make words with each pair so we will use permutation.
We have 26 choices to choose the letter of the word. Now since the repetition is not allowed therefore, we are left with 25 choices of letters to fill the second place of the word. Also, for the last place, we are now left with only 24 choices.
Therefore, in order to get the total number of ways in which a three letter word can be formed, we need to multiply all the choices evaluated above together.
Therefore, the total number of words is given by:-
\[ = 26 \times 25 \times 24\]
\[ = 15600{\text{ words}}\]

15600 three letter words can be formed.

Note: We can also use the direct formula of permutation to solve the given problem.
The formula of permutation is given by:-
\[^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] where n is total number of objects and r is the number of objects to be selected.
  We have to select three letters out of 26 letters and then arrange them without repetition.
Hence using the permutation formula we get:-
\[^{26}{P_3} = \dfrac{{26!}}{{\left( {26 - 3} \right)!}}\]
Solving it further we get:-
\[^{26}{P_3} = \dfrac{{26!}}{{\left( {23} \right)!}}\]
Simplifying it further we get:-
\[^{26}{P_3} = \dfrac{{26 \times 25 \times 24 \times 23!}}{{\left( {23} \right)!}}\]
\[
  { \Rightarrow ^{26}}{P_3} = 26 \times 25 \times 24 \\
  { \Rightarrow ^{26}}{P_3} = 26 \times 25 \times 24 \\
 \]