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# Permutation - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• PERMUTATION AS AN ARRANGEMENT is considered one the most difficult concept.

• 135 Questions around this concept.

## Solve by difficulty

How many different six-letter words can be formed using the letters A, B, C, D, E, F, G without repetition?

In how many distinct ways can 25 students be arranged around a circular table, if 6 students are absent on that day and 10 students went to sports meet?

A group of 8 friends, including 3 siblings, is going to sit in a row for a photo. If the siblings must sit next to each other, in how many different ways can they be arranged?

In how many ways can the digits 0, 1, 2, 3, 4, and 5 be arranged to form a six-digit number, where the last digit must be even?

How many distinct six-digit numbers can be formed using the digits 1, 2, 3, 4, 5, and 6, where the middle digit must be odd?

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A password consists of 5 letters, where each letter can be either a vowel (a, e, i, o, u) or a consonant. In addition, the password must have at least two vowels. How many different passwords are possible?

## Concepts Covered - 1

PERMUTATION AS AN ARRANGEMENT

Permutation basically means the arrangement of things. And when we talk about arrangement then the order becomes important if the things to be arranged are different from each other (when things to be arranged are the same then order doesn’t have any role to play). So in permutations order of objects becomes important.

Arranging n objects in r places (Same as arranging n objects taken r at a time) is equivalent to filling r places from n things.

So the number of ways of arranging n objects taken r at a time = n(n - 1) (n - 2) ... (n - r + 1)

$\\\mathrm{\frac{n(n-1)(n-2)...(n-r+1)(n-r)!}{(n-r)!}=\frac{n!}{(n-r)!} = ^{n}P_{\;r}}$

Where $r \leq n\text{ and }r \in W$

So, the number of ways arranging n different objects taken all at a time = $^{n}P_{\;n} = n!$.

Example: In how many ways can 5 people be seated at 3 places?

Solution: Basically this question is about arranging 5 people at 3 different places

Let’s think that we are given 3 places, so for the first place we have 5 people to choose from, hence this can be done in 5 ways as all 5 are available.

Now for 2nd place we have 4 people to choose from, hence this can be done in 4 ways.

Similarly for 3rd place we have 3 choices.

Since we have to choose for all 3 places, so multiplication rule is applicable, and the total number of ways  5×4×3=120,

This can also be done directly from the notation or formula

$\\\mathrm{^nP_r \;where \;n= 5, r=3, \\ so \; ^5P_3 = \frac{5!}{2!}} \mathrm{=5\times 4 \times 3 = 120}$

Example: Find the number of ways the letters of the word “BIRTHDAY” can be arranged taken all at a time.

Solution: From the above concept directly using the formula $\mathrm{^nP_n}$ we have

$\mathrm{^8P_8=8!=40,320}$

## Study it with Videos

PERMUTATION AS AN ARRANGEMENT

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