Permutations

Permutations: Permutation is the arrangement of objects taken, some or all, at a time. The number of permutations of n different objects taken r at a time, where 0 < r ≤ n and the objects do not repeat, 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 find the number of 5 letter words that can be formed out of the letters I,N,P,U,T.

For a 5-letter word, the 5 vacant spaces can be filled using any of the 5 letters. But the repetition of letters is not allowed in the empty spaces.

The first vacant place has 5 different options, I,N,P,U,T. When we choose one letter for the first blank space, only 4 options are left for the second vacancy. Similarly, only 3 options are available for the third space, 2 options remaining for the fourth space, and only one option is left for the last space.

So, the different number of ways in which the 5 vacant places can be filled =5×4×3×2×1=120

Here, we are counting the different possible arrangements of the letters such as INPUT,INPTU,NPTUI ..., etc. In each word, the arrangement of letters is different from one another.

Such an ordered arrangement is called a permutation of 5 distinct letters when all are taken at a time.

We can also find the number of 4-letter words, the number of 3-letter words or even 2-letter words that can be formed out of the letters of the word INPUT. We can do that with or without meaning, where the repetition of the letters is not allowed.

In such cases, we are considering only 4 or 3 or 2 letters at a time, not all the 5.

Hence, we can say that the permutation is an arrangement of distinct objects in a definite order, taking some objects or all at a time.

The number of permutations of n different objects taken r at a time, where 0 < r ≤ n and the objects do not repeat is n(n – 1)(n – 2)...(n – r + 1). This is denoted by ⁿP_r and defined as

Where 0 ≤ r ≤ n

When we consider all objects at a time, then r = n, and

When we don’t consider any objects, then r=0 and

Arranging no objects is the same as leaving all the objects behind. As we know, there is only one way to do so.

Let us consider another example of word making with letters I,N,P,U,T. If repetition of letters is allowed, then for each vacant space, there will be 5 options available.

Then, the total number of options will be 5×5×5×5×5=5⁵=3125

That is, the number of permutations of n different distinct objects taken r at a time, where repetition is allowed, is

So far, we saw the permutations when all the objects are distinct with or without repetition of objects. But the objects need not always be distinct.

For example, consider the word EXPERT. Here, the letter E occurs twice.

That is, out of the 6 letters, 2 letters are the same. The permutation calculation is different in this case.

The number of permutations of n objects, where p objects are of the same kind and the rest is all different, is given by 

Now, consider the example PARALLELOGRAM.

PARALLELOGRAM has 13 letters in which there are 3 As, 3 Ls, 2 Rs, and 5 other distinct letters. The number of permutations of n objects, where p_1 objects are of one kind, p_2 are of the second kind, ..., p_k are of k^th kind and the rest, if anything left, are of a different kind is given by

Q1. How many distinct words can be formed using the letters of the word COCOMELON that have meaning or no meaning at all?

Ans: The number of permutations of n objects, where p1 objects are of one kind, p_2 are of the second kind, ..., p_k are of k^th kind and the rest, if anything left, are of a different kind is given by n!/(p_1 !p_2 !……p_k !)

COCOMELON has 9 letters in which there are 2 C,3 O, and 4 other distinct letters. Hence, the number of distinct words that can be formed by COCOMELON=9!/2!3! =(9×8×7×6×5×4×3×2×1)/(3×2×1×2×1) =30240

Q2. 6 kids are to be arranged in a line for a photo. But 2 of them are identical twins. How many arrangements would be there if the photographer gave the twins a different colour jacket to wear?

Ans: Let the kids be A,B,C,D,E,E. The kids can occupy any place in a row of 6 vacant spaces. The identical twins can sit together or not. Let x be the number of ways of arranging 6 people.

Now when the photographer gives the twins two coloured jackets to wear, there could be two ways to get the same arrangement.

Now we would have 2x ways of arranging the kids. All the 6 kids are now distinct, and they can be arranged in 6! ways.

⇒2x=6! ⇒x=6!/2=(6×5×4×3×2×1)/2=360

Hence, the number of different arrangements possible if the photographer gave the twins a different colour jacket to wear is 360.

Q3. How many ways can we arrange 5 girls out of 8 in 5 chairs in a straight line in a room?

Ans: There is a total of 8 girls and 5 chairs. We need to arrange 5 girls from those 8 in the 5 chairs. Here, as there are only 5 chairs, we should examine the situation from the view of the chairs, not the girls. We must always consider from the point of view of that object which is fewer in number. The number of ways of arranging 5 out of 8 distinct objects can be done by (_ ^n)P_r =n!/(n-r)!=8!/(8-5)!=8!/3!=6720