How many four-letter computer passwords can be formed using only the symmetric letters (no repetition allowed)?
$\left( {\text{A}} \right){\text{ 7920}}$
$\left( {\text{B}} \right){\text{ 330}}$
\[\left( {\text{C}} \right){\text{ 14640}}\]
\[\left( {\text{D}} \right){\text{ 419430}}\]
Answer
300.6k+ views
Hint: In this problem, we have to find the number of ways for four-letter computer passwords can be formed using only the symmetric letters (no repetition allowed). First, we have to construct symmetric letters in the English alphabet and then we take four letters in that form. Here, we have to use the permutation formula to find the required result. We can easily find out the number of ways to form four-letter computer passwords by substituting the required terms into the permutation formula. Finally, we can get an answer.
Formula used:
General formula for permutation, $_{}^nP_r^{} = \dfrac{{n!}}{{(n - r)!}}$
Complete step by step answer:
The question stated that we use only symmetric letters in English alphabet.
Here the symmetric letters are, A, H, I, M, O, T, U, V, W, X, Y
They are $11$ symmetric letters in English alphabets.
Now we have to calculate that number of ways to make four-letter computer password using only symmetric letters,
That is we can write it as,
$ \Rightarrow {}^{11}{P_4}$
Now we use the formula for permutation, ${}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$
Here $n = 11$ and $r = 4$,
Substitute these values in formula we get,
$ \Rightarrow \dfrac{{11!}}{{(11 - 4)!}}$
Let us subtract the denominator term and also split the numerator factorial, we get
$ \Rightarrow \dfrac{{11 \times 10 \times 9 \times 8 \times7!}}{{7!}}$
Cancel the term we get,
$ \Rightarrow 11 \times 10 \times 9 \times 8$
On multiplying the terms we get,
$ \Rightarrow 7920$
Since, we have to choose \[4\] letters from $11$ symmetric letters in English alphabets.
Thus the maximum number of ways by which four-letter passwords can be formed only by using symmetric letters is $7920$.
Note: We have to remember that, In the case of solving the questions by applying permutation, one has to keep in mind that they have to focus on both selection and arrangement. In a simplified way, we can say that ordering is very much essential in permutation. Permutation depends basically on three conditions, firstly, when elements are not permitted to be repeated, secondly when it is appropriate to repeat elements and thirdly, if the set of elements are not distinct.
Formula used:
General formula for permutation, $_{}^nP_r^{} = \dfrac{{n!}}{{(n - r)!}}$
Complete step by step answer:
The question stated that we use only symmetric letters in English alphabet.
Here the symmetric letters are, A, H, I, M, O, T, U, V, W, X, Y
They are $11$ symmetric letters in English alphabets.
Now we have to calculate that number of ways to make four-letter computer password using only symmetric letters,
That is we can write it as,
$ \Rightarrow {}^{11}{P_4}$
Now we use the formula for permutation, ${}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$
Here $n = 11$ and $r = 4$,
Substitute these values in formula we get,
$ \Rightarrow \dfrac{{11!}}{{(11 - 4)!}}$
Let us subtract the denominator term and also split the numerator factorial, we get
$ \Rightarrow \dfrac{{11 \times 10 \times 9 \times 8 \times7!}}{{7!}}$
Cancel the term we get,
$ \Rightarrow 11 \times 10 \times 9 \times 8$
On multiplying the terms we get,
$ \Rightarrow 7920$
Since, we have to choose \[4\] letters from $11$ symmetric letters in English alphabets.
Thus the maximum number of ways by which four-letter passwords can be formed only by using symmetric letters is $7920$.
Note: We have to remember that, In the case of solving the questions by applying permutation, one has to keep in mind that they have to focus on both selection and arrangement. In a simplified way, we can say that ordering is very much essential in permutation. Permutation depends basically on three conditions, firstly, when elements are not permitted to be repeated, secondly when it is appropriate to repeat elements and thirdly, if the set of elements are not distinct.
Recently Updated Pages
If a parabola whose length of latus rectum is 4a touches class 11 maths JEE_Main

Find the cubic polynomial whose zeroes are 3 5 and class 11 maths JEE_Main

During the sale colour pencils were being sold in -class-11-maths-JEE_Main

A man on the top of a vertical observation tower o-class-11-maths-JEE_Main

In a class of 60 students 25 students play cricket class 11 maths JEE_Main

A regular polygon has 20 sides How many triangles can class 11 maths JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
NCERT Solutions For Class 11 Maths Chapter 6 Permutations And Combinations - 2026-27 Free PDF Download (Login Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

NCERT Solutions For Class 11 Maths Chapter 9 Straight Lines - 2026-27 Free PDF Download (Sign-in Required)

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations - 2026-27 Free PDF Download (Login Required)

Hybridisation in Chemistry – Concept, Types & Applications

