Suppose in our library shelf we want to arrange about $35$ books which are of three different subjects. These can be arranged in such a way that the books on the same subjects are together. When we start arranging, our question is in how many ways can this be arranged?

From $10$ students, a committee of $6$ is to be arranged which contain a President, a Vice President and a secretary. In how many ways this can be arranged? When we try to solve these problems practically, it will be very difficult and time consuming. Hence we use the method of permutation and combinations to make these arrangements. In this section let us see about the meaning of the terms Permutations and Combinations and the method of evaluating them using appropriate formulae.

Permutation vs Combination

What is Permutation?

The different arrangements which can be made out of a given number of things by taking some or all at a time are called permutations.

Examples:

1) What are the two digit numbers that can be formed using the digits $2, 5, 7$.

Solution:

Here out of three digits, we formed two digit numbers, (i.e) we form numbers by taking two digits at a time.

The numbers that can be formed are, $25, 27, 52, 57, 72, 75$.

2) What are the different numbers that can be formed using all the digits, $9, 8$ and $5$.

Solution:

Here we form numbers by taking all at a time. (i. e) we form three digit numbers by taking all the digits at a time.

The numbers that can be formed are $589, 598, 859, 895, 958$ and $985$

NOTE: Hence we observe that in permutations there will be selection and then arrangements.
What is Combination?

Each of the different groups or selections which can be formed by taking some or all of a number of objects irrespective of their arrangements, is called a combination.

Example:

1) John as three pens each of blue, red and green. In how many ways two pens can be selected from the three pens.

Solution:

John as three colored pens each of blue, red and green.

The different ways of selecting two pens are, blue and red ; red and green (or) blue and green.

Therefore there are three ways of selecting two pens from three pens.

2) A bag has an yellow marble, black marble and a blue marble. In how many ways three marbles can be selected?

Solution:

Since the bag contains three marbles each of yellow, black and blue, the selection of thee marbles contain all the three colors yellow, black and blue.

Therefore, the selection can be done only one way.

NOTE: Hence we observe that in combinations there will be only selection (and no arrangement).

Formulas

Factorial: Factorial notation is used to express the product of first $n$ natural numbers.

(i. e) $n!$ = $1.2.3.4 . . . . . . . . . . n$

Example:

$5 !$ = $1.2.3.4.5$ = $120$

$10 !$ = $1.2.3.4.5.6.7.8.9.10$ = $3628800$

Permutations Formula:

Formula 1: The permutations of n objects by taking $r$ at $a$ time is,

$P (n, r)$ = $nPr$ = $n (n - 1) (n - 2)$ . . . . . . . . $(n - r + 1)$

Formula 2: The above permutation can be expressed using factorial notation as follows.

$P (n, r)$ = $nPr$ = $\frac{n!}{(n-r)!}$

Example:

In how many ways three digits numbers can be formed using the digits, $3, 4, 5, 7$ and $9$.

Solution:

We have $5$ digits, $3, 4, 5, 7$ and $9$.

The number of three digit numbers that can be formed is permutation of five things taken three at a time.

(i. e), $5P3$ = $5 ( 5 - 1) (5 - 2)$ [ by Formula $1, nPr$ = $n (n-1) (n-2)$ . . . . . . $(n - r + 1)$]

= $5.4.3$

= $60$

Factorial Method: Using the factorial formula, we have $nPr$ = $\frac{n!}{(n-r)!}$

$P (5, 3)$ = $5P3$

= $\frac{5!}{(5-3)!}$

= $\frac{5.4.3.2.1}{2.1}$

= $5.4.3$

= $60$

There will be $60$ three digit numbers that can be formed.

Combinations Formula:

Formula 1: Combination of $n$ objects by taking $r$ at a time is $nCr$ = $C\ (n, r)$ = $\frac{nPr}{r!}$

Formula 2: Combination of $n$ objects by taking $r$ at a time is $nCr$ = $C\ (n, r)$ = $\frac{n!\;}{(n-r)!\;r!}$

Example:

In how many ways $3$ balls can be selected from a box containing $10$ balls.

Solution:

Since the above situation is only selected we have combination.

Therefore, $C\ (n, r)$ = $C(6, 3)$

= $\frac{nPr}{r!}$

= $\frac{6P3}{3!}$

= $\frac{6.5.4}{3.2.1}$

= $20$

(or) $C\ (6, 3)$ = $\frac{n!}{(n-r)!\;r!}$

= $\frac{6!}{3!\;3!}$

= $\frac{1.2.3.4.5.6}{1.2.3\;.1.2.3}$

= $20$

Examples

Solved Examples

Question 1: How many four letter words, with or without meaning can be formed out of the letters of the word, "MATHEMAGIC", if repetition of letters is not allowed.
Solution:

The word "MATHEMAGIC" contain the letters, M, A, T, H, E, G, I, C which are 8 letters.
Here we need to select four letters and arrange them.
Hence we have selection and arrangement which is permutation.
Therefore the required number of words = Permutations of 8 letters by taking four at a time
= P (8, 4)
= 8 (8 - 1) (8 - 2) (8 - 3)
= 8 . 7. 6. 5
= 1680

Question 2: In how many ways can 10 books be arranged on a shelf so that a particular pair of books shall be
a. always together.
b. never together.
Solution:

a. Since a particular pair of books is always together.
If we keep these together as one pair, then we have to arrange 9 books on the shelf.
This can be done in P (9, 9) ways = 9 ! ways
Since the pair of books together can be arranged in 2 ways,
the total number of ways of arranged the ten books so that a particular pair of books is always together = 2 x 9 !

b. The ten books can be arranged in P (10, 10) ways = 10 ! ways
From (a), the ten books can be arranged in 9! . 2 ways by arranging particular pair of books together.
Therefore, the number of ways of arranging the 10 books so that a particular pair is never together is,
=10 ! - 2 x 9 !
= 10 x 9 ! - 2 x 9!
= (10 - 2) x 9 !
= 8 x 9 !
the number of ways of arranging the 10 books so that a particular pair is never together is = 8 x 9 !

Question 3: How many diagonals are there in an octagon?
Solution:

A polygon of 8 sides has 8 vertices.
By joining any two of these vertices, we obtain either a side or a diagonal of the polygon.
Here we have only a selection (and no arrangement) , hence we have combination.
Number of all straight lines obtained by joining 2 vertices at a time
= C (8 , 2)
= $\frac{8!}{6!\;.2!}$

= $\frac{8.7.6.5.4.3.2.1}{1.2.3.4.5.6.1.2}$

= 28
Since the number of sides = 8,
Number of diagonals of the octagon = 28 - 8
= 20

Practice Problems

Practice Problems

Question 1: In how many ways 7 boys and 5 girls be arranged for a group photograph, if the girls are to sit on chairs in a row and the boys are to stand in a row behind them?
Question 2: Find the number of ways in which the letters of the word, "LAPTOP", can be arranged such that the vowels occupy only even position.
Question 3: How many three digit numbers can be formed with the digits, 3, 4, 5, 6, 7, 8, when the digits may be repeated any number of times in any arrangement.
Question 4: Out of 6 men and 4 women a committee of 54 is to be formed containing at least one woman. In how many ways the committee can be formed.
Question 5: A code word is to consists of two distinct alphabets followed by two distinct numbers between 1 and 5. How may such code words are there?