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Permutations & Combinations

Permutations And Combinations

MODULES

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Basics of Permutations
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Basics of Combinations
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Letters Technique
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Using Blanks
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Blanks with Repetition
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Slotting Technique
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Conditional Permutations
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Special Types
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Circular Permutations
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Derangement
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Binomial Expansion
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Forming Groups
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Identical Elements
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Identical Groups
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Selection in Geometry
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Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Permutations & Combinations 1
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Permutations & Combinations 2
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Permutations & Combinations 3
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Permutations & Combinations 4
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Permutations & Combinations 5
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Permutations & Combinations 6
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PRACTICE

Permutations & Combinations : Level 1
Permutations & Combinations : Level 2
Permutations & Combinations : Level 3
ALL MODULES

CAT 2025 Lesson : Permutations & Combinations - Letters Technique

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2. Permutations Technique

(a) If
n\bm{n}n distinct items are to be arranged in n\bm{n}n distinct positions, it can be done in nPn=n!^{n}P_n = \bm{n!}nPn​=n! ways.

Let's try to find the number of ways in
5\bm{5}5 students can be seated in 5\bm{5}5 seats in a row. While the seats might be similar in the above example, the position of these seats (1st1^{\mathrm{st}}1st from left, 2nd2^{\mathrm{nd}}2nd from left, etc.) make them distinct.

∴\therefore∴ Number of ways 555 students can be arranged in 555 seats in a row =5!=120= \bm{5!} = \bm{120}=5!=120

(b) If
n\bm{n}n distinct items are to be given exactly 111 of n\bm{n}n distinct things, it can be done in nPn=n!\bm{^{n}P_n } = \bm{n!}nPn​=n! ways.

Now, let's try to find the number of ways in which
5\bm{5}5 people can be given 5\bm{5}5 different coins, such that each of them receives 111 coin. We can draw parallels to the above example here. The 555 people are distinct and are like the 555 seats. And, the 555 different coins are like the 555 students.

∴\therefore∴ Number of ways 555 people can be given 555 different coins =5!=120= \bm{5!} = \bm{120}=5!=120

2.1 Permutation similar Elements (Letters Technique)

In a total of
n\bm{n}n elements, if there are a\bm{a}a similar elements of one kind, b\bm{b}b similar elements of a second kind, c\bm{c}c similar elements of a third kind and so on, then the n\bm{n}n elements can be arranged in n!a! b! c! ...\dfrac{n!}{a! \space b! \space c! \space ...}a! b! c! ...n!​ ways.

Note: This is applicable when all
n\bm{n}n elements are to be arranged. If only few of the n\bm{n}n elements are to be arranged, you need to break down and find the arrangements one-by-one. This type of question is covered in Example 23\bm{23}23.

Example 5

A shopkeeper needs to arrange 222 Kit kat bars, 444 Mars bars and 333 Perk bars in a row. In how many different ways can she arrange these?

Solution

There are a total of 999 items and all of them need to be arranged.

Similar Elements =
222 Kit kats, 444 Mars and 333 Perk

Number of permutations
=n!a!b!c!=9!2! 4! 3!= \dfrac{n!}{a!b!c!} = \dfrac{9!}{2! \space 4! \space 3!}=a!b!c!n!​=2! 4! 3!9!​

Simplifying this is not difficult. We first remove the largest factorial from the denominator

=9×8×7×6×52!×3!=63×20=1260= \dfrac{9 \times 8 \times 7 \times 6 \times 5}{2! \times 3!} = 63 \times 20 = 1260 =2!×3!9×8×7×6×5​=63×20=1260

Answer: 1260

Example 6

Joe has 777 coloured bricks. 333 of them are red, 222 of them are blue, 111 is yellow and 111 is white. In how many different ways can all of these 777 bricks be arranged one over the other?

Solution

Total items =7= 7=7
Similar Elements
=3= 3=3 red, 222 blue, 111 yellow and 111 white

Number of permutations
=n!a!b!c!d!=7!3!×2×1×1=504012=420= \dfrac{n!}{a!b!c!d!} = \dfrac{7!}{3!\times2\times1\times1} = \dfrac{5040}{12} = \bm{420}=a!b!c!d!n!​=3!×2×1×17!​=125040​=420

If there is only
111 element of 111 kind, we divide by 1!1!1!. This doesn't change the answer. Therefore, going forward, we will consider similar elements only when they are more than 111. Rewriting the above,

Total Items
=7= 7=7
Similar Elements
=3= 3=3 red and 222 blue

Number of permutations
=n!a!b!=7!3!×2!=504012=420= \dfrac{n!}{a!b!} = \dfrac{7!}{3!\times2!} = \dfrac{5040}{12}= \bm{420}=a!b!n!​=3!×2!7!​=125040​=420

Answer:
420420420

2.1.1 Permuting letters in a word

A word can have
222 or more letters of the same kind. These similar letters are like similar elements explained in the above section {2.2} Permutation involving similar elements.

In an
n\bm{n}n-letter word, if a letter occurs a\bm{a}a times, another letter occurs b\bm{b}b times, another letter occurs c\bm{c}c times and so on, then the

Number of different
n\bm{n}n-letter words that can be formed =n!a!b!c!...= \dfrac{n!}{a!b!c!...}=a!b!c!...n!​

For instance, the letters of the word BET can be arranged in the following
666way BET, BTE, EBT, ETB, TBE, TEB

The letters of the word BEE can be arranged in the following
3\bm{3}3 ways only BEE, EBE, EEB
In the case of BEE, there are 2 Es (similar letters).

∴\therefore∴ Number of ways the letters in word

BET can be arranged
=3!=6= 3! = \bm{6}=3!=6 ways

BEE can be arranged
=3!2!=3= \dfrac{3!}{2!} = \bm{3}=2!3!​=3 ways

Example 7

In how many ways can all the letters in the word APPEAR be used
(I) to form
666-letter words?
(II) to form
666-letter words that begin with A and end with R?

Solution

Case I: In the 666-letter word APPEAR, there are 222 As and 222 Ps.

∴\therefore∴ Number of ways to form 666-letter words =6!2!×2!== \dfrac{6!}{2!\times2!} ==2!×2!6!​= 180\bm{180}180 ways

Case II: If A and R are fixed, then we will be able to permute only the other
444 letters that are PPEA.

∴\therefore∴ Number of ways to form words using PPEA =4!2!=12= \dfrac{4!}{2!} = \bm{12}=2!4!​=12 ways

Answer: (I)
180180180 ways; (II) 121212 ways

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