Factorial is defined as the product of all natural numbers less than or equal to a given natural number.

For a natural number ‘n’, its factorial is denoted by n!

And, n! = n (n-1) (n-2) (n-3) (n-4) …... 3 x 2 x 1.

For example: 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040

10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3628800

Note: (1) 0! = 1

(2) 1! = 1

The arrangement of objects in a definite order is called Permutation.

For example: You have two books, one book of each subject, Mathematics and Science. You want to keep them properly on a shelf. So, you can keep them either Mathematics book first and then science book (i.e. M S), or Science book first and then Mathematics book next to it (i.e. S M). Therefore, there are two ways to arrange the two books on a shelf.

The number of permutations of n different objects taken r objects out of them without replacement, where 0 < r ≤ n, is given by:

nPr = \[\frac{n!}{(n-r)!}\]

From previous example, the number of permutations of 2 different books, Mathematics and Science, taken both of them is:

2P2 = \[\frac{2!}{(2-2)!}\] = \[\frac{2!}{0!}\] = 2! = 2 x 1 = 2 ways

The selection of some or all objects from a given set of different objects where order of selection is not considered is called Combination.

For example: Suppose you have three friends, Aman, Mohit and Raj, you want two of them to go with you for a picnic. You are not able to decide which two of them you should take for picnic, so you think of an idea that you will write each of their names on a separate paper and pick two of them. Then, the possible ways you could get the names of your two friends will be:

Aman and Mohit

Mohit and Raj

Aman and Raj

Therefore, there are three ways to select two friends out of three friends.

The number of combinations of n different objects taken r objects out of them without replacement, where 0 < r ≤ n, is given by:

nCr = \[\frac{n!}{r!(n-r)!}\] = \[\frac{ₙPr}{r!}\]

From previous example, the number of ways of picking the names of two friends out of three names is:

3C2 = \[\frac{3!}{2!(3-2)!}\]= \[\frac{3!}{2! Х 1!}\] = \[\frac{3Х2Х1}{2Х1}\] = 3 Ways

Note:

(1) If in a problem statement, you are asked for selection and their ordering, then you should use Permutation.

In simple words, Permutation = Selection + ordering

(2) If in a problem statement, you are asked only for selection then you should use Combination.

In simple words, Combination = Selection

Q.1. How many words can be formed from the letters of the word ‘VEDANTU’ using each letter only once?

Solution: There are 7 letters in the word ‘VEDANTU’.

The number of ways of arranging all the 7 letters of given word is:

7P7 = \[\frac{7!}{(7-7)!}\] = \[\frac{7!}{0!}\] = 7! = 5040 words

Q.2. How many 5digit telephone numbers can be formed if each number starts with 21 and no digit appears more than once?

Solution: Since, the first two places have to be filled only by 2 and 1 respectively and there is only 1 way for doing this.

Also, it is given that no digit appears more than once. Hence, there will be 8 digits remaining (0,3,4,5,6,7,8,9) and 3 places have to be filled with these remaining digits.

So, the next 3 places can be filled with the remaining 8 digits in 8P3 ways.

Total number of ways = 8P3 = \[\frac{8!}{(8-3)!}\] = \[\frac{8!}{5!}\] = \[\frac{8Х7Х6Х5Х4Х3Х2Х1}{5Х4Х3Х2Х1}\] = 8×7×6 = 336

Q.3. How many chords can be drawn through 15 points on a circle?

Solution: Since, a chord is drawn by joining two points on a circle.

Given that, there are 15 points on a circle. The total number of ways of selecting two points on a circle will give the total number of chords of the circle.

Therefore, the required number of chords = 15C2 = \[\frac{15!}{2!(15-2)!}\] = \[\frac{15!}{2!(13)!}\] = \[\frac{15Х14}{2Х1}\] = 105