A permutation is a mathematical technique that determines the number of possible arrangements in a set when the order of the arrangements matters. Let us understand this concept using the below example:
Stacy has 3 dresses and 2 handbags. There are three ways in which a dress can be chosen as there are 3 dresses available. For every choice of a dress, there are two choices of the handbag. Therefore, there are 3 × 2 = 6 pairs of a dress and a handbag. Let 3 dresses be represented as D1, D2, and D3 and the two handbags as H1 and H2.
The fundamental principle of counting states that-
| If an event can occur in m different ways, following which another event can occur in n different ways then the total number of occurrences of the event in the given order is m × n. |
similarly, for three events, the principle is as follows:
If an event can occur in m different ways, following which another event occurs in n different ways, following which a third event can occur in p different ways, then the total number of occurrences of the event in the given order is m × n × p
In our example, the different ways of choosing a dress and a handbag were the number of different ways of the occurrence of the following event in succession:
Example 1: Given 4 different colors of light bulbs red, green, blue, and yellow. How many different signals can be generated if a signal requires the use of two bulbs one below the other?
Solution: There are two vacant places to fill up with these four different colored bulbs.
The upper vacant place can be occupied by any of the four bulbs, hence the top place can be filled up in 4 different ways. The lower vacant place can be filled by the remaining 3 bulbs, hence the lower space can be filled up in 3 different ways. Therefore, the required number of signals that we can generate using 4 bulbs are 4 × 3 =12.
Here we are counting the permutations of 4 different bulbs taken 2 at a time.
The product of first n natural numbers 1 × 2 × 3 × 4 × ... × (n−1) × n is called 'n factorial' and is denoted by \(n!\)
Remember \(0! = 1 \)
Example 2: Evaluate \(5!\)
Solution: 5 × 4 × 3 × 2 × 1 = 120
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The number of permutations of n different objects taken r at a time is denoted by \(^nP_r\). |
| The number of permutations of n different objects taken r at a time where repetition is allowed is \(n^r\) |
Let us understand both of these theorems using an example.
Example 3: Find the number of 4 letter words with or without meaning, which can be formed out of the letters of the word HEAD, where the repetition of letters is not allowed.
Solution: Using theorem 1, \(^nP_r = \frac{n!}{(n-r)!}\), where n is 4 and r is 4, therefore
\(^4 P _4 = \frac{4!}{(4-4)!} = \frac{4!}{1} = 24\)
Now consider if the repetition of letters are allowed, then the number of permutations possible as per theorem 2 are \(4^4 = 256\).
Example 4: Find the number of ways a principal and vice-principal can be chosen from a group of 10 people such that the same person cannot hold both the position.
Solution: \(^{10} P _2 = \frac{10!}{(10-2)!} = \frac{10!}{8!} = 10 \times 9 = 90\), 10 people taken 2 at a time
| The number of permutations of n objects, where p1 objects are of the same kind, p2 are of the second kind,... pk are of kth kind and the rest, if any, are of a different kind is \(\frac{n!}{p_1!p_2!...p_k!} \) |
Let us understand the application of this theorem using the below example
Example 5: Find the number of arrangements of the letters of the word INDEPENDENCE.
Solution: As the letters are repeating, so we use the formula \(\frac{n!}{p_1!p_2!...p_k!} \). There are 3 N, 4 E, 2 D, 1 I, 1 P and 1 C.
\(\mathcal{\frac{12!}{3!4!2!} }\)= 1663200
Let us change the above question as: find the number of arrangements where the word start with P
Here as the position of letter P is fixed so count the arrangement of the remaining 11 letters.
\(\frac{11!}{3!4!2!} = 138600\)
