Chapter 6: Permutations and Combinations

6.1 Fundamental Principle of Counting

“If an event can occur in \( m \) different ways, following which another event can occur in \( n \) different ways, then the total number of occurrence of the events in the given order is \( m \times n \).” This is known as the Multiplication Principle.

Alternatively, the Addition Principle: “If there are two jobs such that they can be performed independently in \( m \) and \( n \) ways respectively, then either of the two jobs can be performed in \( (m + n) \) ways.”

6.2 Factorial Notation

The notation \( n! \) represents the product of first \( n \) natural numbers. \[ n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1 \] By convention, \( 0! = 1 \).

6.3 Permutations

A permutation is an arrangement in a definite order of a number of objects taken some or all at a time. The number of permutations of \( n \) different objects taken \( r \) at a time, where \( 0 \le r \le n \), is denoted by \( {}^nP_r \) and is given by: \[ {}^nP_r = \frac{n!}{(n - r)!} \]

Permutations when all objects are not distinct

The number of permutations of \( n \) objects, where \( p_1 \) objects are of one kind, \( p_2 \) are of second kind, …, \( p_k \) are of \( k^{\text{th}} \) kind and the rest, if any, are of different kind is: \[ \frac{n!}{p_1! p_2! \dots p_k!} \]

6.4 Combinations

A combination is a selection of some or all of a number of different objects where the order of selection is ignored. The number of combinations of \( n \) different objects taken \( r \) at a time, where \( 0 \le r \le n \), is denoted by \( {}^nC_r \) and is given by: \[ {}^nC_r = \frac{n!}{r!(n - r)!} \]

Important Properties

• \( {}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r \)
• \( {}^nC_r = {}^nC_{n-r} \) (Choosing \( r \) elements is identical to rejecting \( n-r \) elements)

Competency-Based Questions

Question 1:

A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of exactly 1 man and 2 women?

Answer 1:

Total number of ways to form the committee:

Since we are selecting 3 persons out of 5 (2 men + 3 women) and order does not matter, this is a combination problem.

The total number of ways = \( {}^5C_3 \).

\[ {}^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = \frac{20}{2} = 10 \]

So, there are 10 ways to form the committee.

Committees with exactly 1 man and 2 women:

We need to select 1 man from 2 men AND 2 women from 3 women.

By the Fundamental Principle of Counting, we multiply the combinations:

Number of ways = \( {}^2C_1 \times {}^3C_2 \)

\[ {}^2C_1 = 2 \]

\[ {}^3C_2 = \frac{3!}{2!1!} = 3 \]

Total = \( 2 \times 3 = 6 \)

Conclusion: There are 6 committees that consists of exactly 1 man and 2 women out of total 10 committees.

Question 2:

How many words, with or without meaning, can be formed using all the letters of the word “EQUATION” at a time so that the vowels and consonants occur together?

Answer 2:

The word “EQUATION” has 8 letters.

Vowels (V): E, U, A, I, O (5 vowels)

Consonants (C): Q, T, N (3 consonants)

Since vowels and consonants must occur together, we can block them into two groups: [V] and [C].

These 2 groups can be arranged among themselves in \( 2! \) ways.

Inside group [V], the 5 vowels can be arranged in \( 5! \) ways.

Inside group [C], the 3 consonants can be arranged in \( 3! \) ways.

By the Multiplication Principle, the total number of permutations is:

\[ 2! \times 5! \times 3! = 2 \times 120 \times 6 = 1440 \]

Conclusion: 1440 different words can be formed.