CHAPTER 2 - Introduction to Permutations

The fundamental issue in Permutations is the arrangement of things. In the last example, we had  letters and  places where we could arrange ( of) those letters, We calculated that there are  ways of arranging those  letters, taken  at a time. In mathematical terminology, we calculated as  the number of permutations (arrangements) of  letters taken  at a time.

Let us generalize this: Suppose we have  people. It we had  seats available to seat these people, the total number of ways to do so would be (by the logic discussed in the preceding section) . This quantity is denoted by !.

Suppose now that we have only  seats, where . The total number of ways now would be . This is the number of ways of permuting  things, taken  at a time, and the notation used for this number is . Thus :

Thus, using the fundamental principle of counting, you see that we’ve been able to calculate the value of .

Finally, note that 

Lets apply this discussion to an example:

Example: 1

Find the number of permutations of the word  which (a) consist of all the letters of this word (b) start with  and end with  (c) contain the string “” (d) have the letters and  occurring together (e) start and end with vowels (f) have no two vowels occurring together.

Solution: 1-(a)

We have  letters and we want to permute all of them. The required number of arrangements would be

Solution: 1-(b)

Fix  at the start and  at the end of the word We now have  places which we need to fill using the remaining  letters. The number of such permutations will be .

Solution: 1-(c)

We want all those permutations in which the string “” occurs. Let us treat “” as a single letter/object since this is what we want – we want “” to appear as a single entity. We now have the following objects (letters) which we need to permute:   “”  These are in total  objects. Thus, they can be permuted in  ways. This is the number of permutations that contain the string ““.

Solution: 1-(d)

We now need the permutations which contain the letters  and occurring together. This means that they can occur together in any order. There are  ways in which  and  can be arranged among themselves, namely  and . Now consider from the previous part all those permutations in which the string ““occurs. These are  in number. Corresponding to each such permutation, we can have  permutations in which the constraint is that  and  occur together. This is because the string “” can itself be permuted in  ways as described above. For example, consider the permutation ““; to this permutation will correspond permutations in which  and  occur together: The required number of permutations is therefore .

Solution: 1-(e)

In the word , there are a total of  vowels. We want to have the starting and ending letters of our permutations as vowels. Let us first arrange the starting and ending letters and then fill in the rest of the places. Out of the  vowels available to us, the first and the last place can be filled in  places. Once we’ve fixed the first and the last letters, the remaining  places can be filled in  ways. Thus, the total number of required permutations is .

Solution: 1-(f)

There are  vowels and  consonants in the word . We require permutations in which no two vowels occur together. We can ensure such permutations by first fixing the  consonants and then arranging the vowels in the  possible places that arise as depicted in the following figure. Convince yourself that any arrangement of the  vowels in the  blank spaces above will correspond to a permutation with no two vowels together. The number of ways of arranging the  consonants is . After the consonants have been arranged, the number of ways of arranging the vowels in the  blank spaces as depicted in Fig. is . Thus, the required number of permutations is

It would be a good exercise for you to construct more of such examples on your own and solve them. You can take an arbitrary word and find the number of permutations of that word, that satisfy any particular condition that you can think of.