How many words can be formed by arranging the letters of the word arrangement so that the vowels remains together?

Question:

How many words can be formed by arranging the letters of the word ‘ARRANGEMENT’, so that the vowels remain together?

Solution:

To find: number of words where vowels are together

Vowels in the above word are: $A, A, E, E$

Consonants in the above word: R,R,N,G,M,N,T

Let us denote the all the vowels by a single letter say $Z$

$\Rightarrow$ The word now has the letters, R,R,N,G,M,N,T,Z

$\mathrm{R}$ and $\mathrm{N}$ are repeated twice

Number of permutations $=\frac{8 !}{2 ! 2 !}$

Now $Z$ is comprised of 4 letters which can be permuted amongst themselves

A and E are repeated twice

$\Rightarrow$ Number of permutations of $Z=\frac{4 !}{2 ! 2 !}$

$\Rightarrow$ Total number of permutations $=\frac{8 ! \times 4 !}{2 !^{4}}=60480$

The number of words that can be formed is 60480

How many words can be formed by arranging the letters of the word ‘ARRANGEMENT’, so that the vowels remain together?

The word ARRANGEMENT has $11$ letters, not all of them distinct. Imagine that they are written on little Scrabble squares. And suppose we have $11$ consecutive slots into which to put these squares.

There are $\dbinom{11}{2}$ ways to choose the slots where the two A's will go. For each of these ways, there are $\dbinom{9}{2}$ ways to decide where the two R's will go. For every decision about the A's and R's, there are $\dbinom{7}{2}$ ways to decide where the N's will go. Similarly, there are now $\dbinom{5}{2}$ ways to decide where the E's will go. That leaves $3$ gaps, and $3$ singleton letters, which can be arranged in $3!$ ways, for a total of $$\binom{11}{2}\binom{9}{2}\binom{7}{2}\binom{5}{2}3!.$$