Mathematics > Permutations and Combinations

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

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  and the objects do not repeat is , which is denoted by n.

Factorial Notation: The notation n! Represents the product of first n natural numbers, i.e., the product  is denoted as

• n
• The number of permutations of n objects, where p objects are of the same kind and rest are all different is
• The number of permutations of n objects, where objects are of one kind,  are of second kind,  are of kth kind and the rest, if any, are of different kind is

Combinations

•          nPr = nCr r!
• n =
• n = n
• If n  = n

Examples

Question

How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?

Solution

There will be as many ways as there are ways of filling 2 vacant places in succession by the five given digits. Here, in this case, we start filling in unit's place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten's place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2 × 5, i.e., 10.

Question

Find the number of permutations of the letters of the word ALLAHABAD.

Solution

Here, there are 9 objects (letters) of which there are 4A's, 2 L's and rest are all different.

Therefore, the required number of arrangements =

Question

In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable?

Solution

Total numbers of discs are 4 + 3 + 2 = 9. Out of 9 discs, 4 are of the first kind (red), 3 are of the second kind (yellow) and 2 are of  the third kind (green).

Therefore, the number of arrangements

Question

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 1 man and 2 women?

Solution

Here, order does not matter. Therefore, we need to count combinations. There will be as many committees as there are combinations of 5 different persons taken 3 at a time. Hence, the required number of ways = 5C3 = 10.

Now, 1 man can be selected from 2 men in 2C1 ways and 2 women can be selected from 3 women in 3C2 ways.

Therefore, the required number of committees = 2C1 * 3C2 = 6.