You are playing a video game and you have reached a stage where to unlock the next level you need to enter a 5 letter word as the passcode. You have collected the clues from the previous levels which are 5 letters M, E, S, I and L. Now, what are the possible ways in which you can frame the passcode? The passcode could be ** LIMES, MILES, SLIME, SMILE**, etc.What is important here is the order of writing the words. Every different possible arrangement of a certain number of items taken all or some at a time is known as permutations.

Now, the total number of possible passcodes would be 5 × 4 × 3 × 2 × 1 = 120 since the letters cannot be repeated. If repetition was possible then the total number of passcodes that could be formed would be 5 × 5 × 5 × 5 × 5 = 3125. One of the examples explaining permutations effectively are the anagrams. In anagrams, the arrangement of letters is crucial in order to make out a meaningful word. As mentioned in the examples above*, *** LIMES, MILES, SLIME, SMILE** are all anagrams.

Applications of permutations and combinations are just not limited to alphabets or words, it can be applied to almost every situation where an arrangement or selection is possible. Depending upon the arrangement and selection, multiple cases may arise for possible ways of arrangement.

**Permutations**

Let us discuss the different cases which would help us deal with problems related to permutations more effectively.

__Permutation when all the objects are distinct and repetition is not allowed__

If n is a positive integer and r is a whole number such that r <n, then P(n, r) represents the number of all possible arrangements or permutations of n distinct objects taken r at a time. It can also be represented as ^{n}P_{r}.

P(n, r) = n(n-1)(n-2)(n-3)……..upto r factors

⇒P(n, r) = n(n-1)(n-2)(n-3)……..(n – r +1)

**Example:** How many 3 letter words with or without meaning can be formed out of the letters of the word ** SWING **when repetition of letters is not allowed?

**Solution:** Here n = 5, as the word SWING has 5 letters. Since we have to frame 3 letter words with or without meaning and without repetition, therefore total permutations possible are:

__Permutation when all the objects are distinct and repetition is allowed__

The number of ways in which n objects can be arranged taken r at a time if repetition of objects is allowed is given by n^{r}.

**Example:** How many 3 letter words with or without meaning can be formed out of the letters of the word ** SMOKE** when repetition of words is allowed?

**Solution:** The number of permutations possible in this case is n^{r} i.e. 5^{3}.

- The number of permutations of n different objects when p
_{1}objects among the ‘n’ objects are alike, p_{2}objects of the second kind are alike, p_{3}objects of the third kind are alike and …….. p_{k}objects of k^{th}kind are alike and the remaining are of a different kind is given by:

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