Permutations and Combinations
Fundamental Principle of Counting: If an event occurs in m different ways, following
which another event occurs in n different ways, then the total number of occurrence of the
events in the given order is m × n. This is called the fundamental principle of counting.
Example:Find the number of 5-letter words, with or without meaning, which can be
formed out of the letters of the word MATHS, where the repetition of digits is not allowed.
Solution: There are as many words as there are ways of filling 5 vacant
places by the 5 letters.
The first place can be filled with any of the 5 letters in 5 different ways, following which the
second place can be filled with any of the remaining 4 letters in 4 different ways, following
which the third place can be filled in 3 different ways, following which the fourth place
can be in 2 different ways, following which the fifth place can be filled in 1 way. Thus, the
number of ways in which the 5 places can be filled, by the multiplication principle, is 5 × 4 ×
3 × 2 × 1 = 120.
Note: If repetition of letters had been allowed, then the required number of words would
be 5 × 5 × 5 × 5 × 5 = 3125.
Factorial notation: The notation n! represents the product of the first n natural numbers,
i.e. n! = n × (n − 1) × (n − 2) × …… × 5 × 4 × 3 × 2 × 1
0! = 1
Example:
Permutation when all objects are distinct: 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 things taken r at a time, when
o repetition is not allowed, is nPr = , where 0 ≤ r ≤ n.
o repetition is allowed, is nr , where 0 ≤ r ≤ n.
Example 1: Twenty five students are participating in a competition. In how many ways,
can the first three prizes be won in such a way that a prize cannot be shared by more than
one student?
Solution: The total number of ways in which first three prizes can be won is the number of
arrangements of 25 different things taken 3 at a time.
So, required number of ways = 25P3
Example 2: Find the total number of four digit numbers that can be formed by using the
digits 0, 2, 5, and 6?
Solution: A four digit number has four places i.e., units, tens, hundreds and thousands.
Units, tens and hundreds place can be filled with either 0, 2, 5, or 6 where as thousands
place can be filled with 2, 5 or 6 only.
Number of ways to fill the units place = 4
Number of ways to fill the tens place = 4
Number of ways to fill the hundreds place = 4
Number of ways to fill the thousands place = 3 ways.
∴ Total number of four digit numbers = 4 × 4 × 4 × 3 = 192
Concept of permutations when all objects are not distinct
o The number of permutations of n objects, when p objects are of the same kind and the rest
are all different, is .
o In general, the number of permutations of n objects, when p1 objects are of one kind, p2 are
of the second kind , … , pk are of the kth kind and the rest, if any, are of different kinds,
is .
Example: Find the number of permutations of the letters of the word ARRANGEMENT.
Solution: Here, there are 11 objects (letters) of which there are 2A's, 2R's, 2N's, 2E's and
the rest are all different.
∴ Required number of arrangements
Combinations: The number of combinations of n different things taken r at a time is
denoted by nCr , which is given by
In particular, nC0 = nCn = 1
Example 1: A box contains 8 red bulbs and 5 blue bulbs. Determine the number of ways in
which 4 red and 2 blue bulbs can be selected.
Solution: It is given that a box contains 8 red bulbs and 5 blue bulbs.
Now, 4 red bulbs can be selected from 8 red bulbs in 8C4 number of ways, and 2 blue bulbs
can be selected from 4 blue bulbs in 4C2 number of ways.
Hence, 4 red bulbs and 2 blue bulbs can be selected from a box containing 8 red bulbs and 4
blue bulbs in 8C4 × 4C2 number of ways.
Now,
8C4 × 4C2
Thus, the number of ways of selecting the bulbs is 420.
In other words, selecting r objects out of n objects is the same as rejecting (n – r) objects.
nCa = nCb ⇒ a = b or a = n, i.e., n = a + b
Example 2: If 19C3r = 19C2r + 4, then find the value of r.
Solution:
19C3r = 19C2r + 4
⇒ 3r + (2r + 4) = 19 or 3r = 2r + 4
⇒ 3r + (2r + 4) = 19 ⇒ r=4
⇒ 5r + 4 = 19
⇒ 5r = 19 – 4 = 15
⇒r=3
∴ The value of r is either 3 or 4.
nCr – 1 + nCr = n + 1Cr
Example 3:If nCr – 1 + nCr + n + 1Cr + 1 + n + 2Cr + 2 = n + aCr + (a – 1), then find the value of a.
Solution:
nCr – 1 + nCr + n + 1Cr + 1 + n + 2Cr + 2
= n +1Cr + n + 1Cr + 1 + n + 2Cr + 2
= n + 1Cr + 1 + n + 2Cr + 2
= n + 3Cr + 2
∴ n + 3Cr + 2 = n + aCr + (a – 1)
⇒a=3
