Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How many 3 letter codes can be formed Using the first 10 letters of English alphabet, if no letter can be repeated?

seo-qna
Last updated date: 13th Jul 2024
Total views: 346.2k
Views today: 4.46k
Answer
VerifiedVerified
346.2k+ views
Hint: Here the given question is based on the concept of permutation. Here we have to find how many 3 letter codes can be formed using the first 10 letters of English alphabet and the repetition of letters is not allowed. We use permutation concepts and we determine the solution for the question.

Complete step-by-step solution:
In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word "permutation" also refers to the act or process of changing the linear order of an ordered set.
Now consider the given question,
 Here we have the first 10 letters of English alphabet, and need to find the number of ways where the 3 letters code can be formed without repeating any letters.
There are 10 letters of English alphabet.
As we know that repetition of the letters is not allowed,
so, the first place can be filled by any of 10 letters.
 Second place can be filled with any of the remaining 9 letters.
The third place can be filled with any of the remaining 8 letters.
Therefore, Number of 3-letter code can be formed when the repetition of letters is not allowed
\[ \Rightarrow \,\,\,10 \times 9 \times 8\]
On multiplication, we get
\[\therefore \,\,\,720\,\,ways\]
Therefore a total of 720 ways is possible.

Note: We can also find these type of question by directly using the permutation formula i.e., \[^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], Here, “\[^n{P_r}\]” represents the “r” objects to be selected from “n” objects without repetition, in which the order matters.
In the given question we have taken 3 letters to be selected from 10 letters without repetition, then by formula
\[ \Rightarrow \,\,{\,^{10}}{P_3} = \dfrac{{10!}}{{\left( {10 - 3} \right)!}}\]
\[ \Rightarrow \dfrac{{10!}}{{7!}}\]
\[ \Rightarrow \dfrac{{10 \times 9 \times 8 \times 7!}}{{7!}}\]
On simplification, we get
\[\therefore \,\,\,720\,ways\]