CAT 2025 Lesson : Permutations & Combinations - Letters Technique

2. Permutations Technique

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

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

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

(b) If $\bm{n}$ distinct items are to be given exactly $1$ of $\bm{n}$ distinct things, it can be done in $\bm{^{n}P_n } = \bm{n!}$ ways.

Now, let's try to find the number of ways in which $\bm{5}$ people can be given $\bm{5}$ different coins, such that each of them receives $1$ coin. We can draw parallels to the above example here. The $5$ people are distinct and are like the $5$ seats. And, the $5$ different coins are like the $5$ students.

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

2.1 Permutation similar Elements (Letters Technique)

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

Note: This is applicable when all $\bm{n}$ elements are to be arranged. If only few of the $\bm{n}$ 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 $\bm{23}$.

2.1.1 Permuting letters in a word

A word can have $2$ 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 $\bm{n}$-letter word, if a letter occurs $\bm{a}$ times, another letter occurs $\bm{b}$ times, another letter occurs $\bm{c}$ times and so on, then the

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

For instance, the letters of the word BET can be arranged in the following $6$way BET, BTE, EBT, ETB, TBE, TEB

The letters of the word BEE can be arranged in the following $\bm{3}$ 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 $= 3! = \bm{6}$ ways

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