Aptitude Day 113

Problems on Permutations and Combinations:

1. How many words can be formed by using all letters of the word 'TIME' so that the vowels are never together?

Solution:
                The word contains 4 different letters. When the vowels always come together, we may treat that whole entity as 1 letter.

                The letters to be arranged are TM (IE).

                These 3 letters can be arranged in ³P3 = 3! = 6 ways.

                The vowels in (IE) may be arranged in 2! = 2 ways.

                The number of words each having vowels together = 6*2 =  12.

                The total number of words formed by using all the letters of the given word = 4! = 24.

                The number of words each having vowels never together = 24-12 = 12.


2.  How many words can be formed by using all letters of the word 'FLUTTER' so that the vowels are always together?

Solution:
                The word has 7 letters in which T occurs twice and others are different. So the arrangement will be FLTTR (UE).

                 The number of ways of arranging these letters = 6!/2! = 360.
                 [ Note: Here 2! is used for the repeatation]

                The vowels may be arranged in 2! = 2.

                The required number of ways = 360*2 =  720.