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Permutations & Combinations
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CAT 2025 Lesson : Permutations & Combinations - Letters Technique

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2. Permutations Technique

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

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

∴\therefore Number of ways 55 students can be arranged in 55 seats in a row =5!=120= \bm{5!} = \bm{120}

(b) If n\bm{n} distinct items are to be given exactly 11 of n\bm{n} distinct things, it can be done in nPn=n!\bm{^{n}P_n } = \bm{n!} ways.

Now, let's try to find the number of ways in which 5\bm{5} people can be given 5\bm{5} different coins, such that each of them receives 11 coin. We can draw parallels to the above example here. The 55 people are distinct and are like the 55 seats. And, the 55 different coins are like the 55 students.

∴\therefore Number of ways 55 people can be given 55 different coins =5!=120= \bm{5!} = \bm{120}

2.1 Permutation similar Elements (Letters Technique)

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

Note: This is applicable when all n\bm{n} elements are to be arranged. If only few of the n\bm{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}.

Example 5

A shopkeeper needs to arrange 22 Kit kat bars, 44 Mars bars and 33 Perk bars in a row. In how many different ways can she arrange these?

Solution

There are a total of 99 items and all of them need to be arranged.

Similar Elements = 22 Kit kats, 44 Mars and 33 Perk

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

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

Answer: 1260

Example 6

Joe has 77 coloured bricks. 33 of them are red, 22 of them are blue, 11 is yellow and 11 is white. In how many different ways can all of these 77 bricks be arranged one over the other?

Solution

Total items =7= 7
Similar Elements =3= 3 red, 22 blue, 11 yellow and 11 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}

If there is only 11 element of 11 kind, we divide by 1!1!. This doesn't change the answer. Therefore, going forward, we will consider similar elements only when they are more than 11. Rewriting the above,

Total Items =7= 7
Similar Elements =3= 3 red and 22 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}

Answer: 420420

2.1.1 Permuting letters in a word

A word can have 22 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}-letter word, if a letter occurs a\bm{a} times, another letter occurs b\bm{b} times, another letter occurs c\bm{c} times and so on, then the

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

For instance, the letters of the word BET can be arranged in the following 66way BET, BTE, EBT, ETB, TBE, TEB

The letters of the word BEE can be arranged in the following 3\bm{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} ways

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

Example 7

In how many ways can all the letters in the word APPEAR be used
(I) to form 66-letter words?
(II) to form 66-letter words that begin with A and end with R?

Solution

Case I: In the 66-letter word APPEAR, there are 22 As and 22 Ps.

∴\therefore Number of ways to form 66-letter words =6!2!×2!== \dfrac{6!}{2!\times2!} = 180\bm{180} ways

Case II: If A and R are fixed, then we will be able to permute only the other 44 letters that are PPEA.

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

Answer: (I) 180180 ways; (II) 1212 ways

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