Question

How many words can be made out of the letters of the word INDEPENDENCE in which the vowels always come together?
A) 16800
B) 16630
C) 1663200
D) None of these

Answer 1

In this type of problem we need to use concept of permutation. We know that a permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement, and use formula for total arrangement, multiple types of questions related to permutation can be solved. One could say it is an ordered combination.

Complete Step-by-step Solution
There are 12 letters in the word “Independence” with 5 vowels ad 7 consonants. Now, bunch up all vowels together as one letter so as to assume a word with 3 letters which can be arranged in :
$= \dfrac{{3!}}{{3!.2!}} = 3360{\text{ ways}}$.
If all the letters were different, the number of words would be :
$12! = 479001600$.
Counting repeatedly permutations of E’s, N’s and D’s,
Therefore, the correct number of words is:
${N_1} = \dfrac{{12!}}{{4!.3!.2!}} \\\ = \dfrac{{479001600}}{{24.6.2}} \\\ = 1663200 \\\$
If the vowels have to be together, first we have to calculate how many are there (same technique)and bunch of vowels can be arranged among themselves in:
${N_2} = \dfrac{{5!}}{{4!}} = 5$ways
Obtained- permutate 8 objects, 7 consonants and 1 vowel, but there are 3 N’s and 2 D’s, so total arrangement of the letters with vowels always coming together is:
${N_3} = \dfrac{{8!}}{{3!.2!}}.{N_2} = 3360.5 = 16800$

$\therefore$ Option (B) is the correct option.

Note:
In the number of permutations of the word “Independence”, the set of vowels themselves form a multiset of 5 letters with the letter e being repeated five times. We need be careful at calculation steps and while writing factorials.