Word formation is based on fundamental principle of multiplication. For example by using the letters A₁, A₂, ……A_n, without repetition n! words can be formed and if A₁, A₂,……. A_r are identical then total number of words without repetition will be \(\cfrac{{n!}}{{r!}}\). For example total number of words which can be formed by using the letters of the word “ACHILLES” will be \(\cfrac{{8!}}{{2!}}\), as the letter L appeared  twice. Let us discuss the following example:

Example: By using A, B, C, D and E how many words can be formed without repetition such that:

(a) All the words start with A.

(b) All the words start with A and end with E

(c) A and B are always together.

(d) A and B are never together.

Solution: (a)   First placed is fixed as A, then there are only 4 letters which can be placed on four places.

__ __ __ __ __

A 4 3 2 1

Total number of words = 4.3.2.1 = 24.

(b) Now first and last both places are fixed.

__ __ __ __ __

A 3 2 1 E

Total number of words = 3.2.1 = 6

(c) If we fix A & B together, so that none of A & B can be separated. So pair A & B will behave   like a single letter.  Let the block AB = x

Now total number of letters are 4 letters ×, C, D, E where × can be AB or BA. So total number of words = (4!). (2!) = 48

(d) Required number of words can be given by:

Total number of words – those when A&B are together. Hence the answer is 5! – (4!) (2!) = 72