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