Permutations approach

Question

There are two questions:

Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels do not occur together.

The solution to this was taken as $8!-6!\times 3!$

In the question:

Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements do all the vowels always occur together?

The solution is $\frac{8!}{3!2!} \times \frac{5!}{4!}$

From where and why did $\frac{5!}{4!}$ come from?

Why is my solution: $\frac{8!}{3!2!} - \frac{5!}{4!}$ wrong?

What am I doing wrong here? How should I approach a permutation question?

Answer 1

Solution

If the five vowels $I, E, E, E, E$ always occur together, they form a block. Thus, we have eight objects to arrange, three $N$s, two $D$s, one $P$, one $C$, and the block of vowels. We can choose the positions of the three $N$s in $\binom{8}{3}$ ways, choose two of the remaining six positions for the two $D$s in $\binom{5}{2}$ ways, and arrange the three distinct objects (the $P$, $C$, and block of vowels) in the remaining three positions in $3$ ways. We can choose four positions for the $E$s within the block in $\binom{5}{4}$ ways. Hence, the number of admissible arrangements is $$\binom{8}{3}\binom{5}{2}3! \cdot \binom{5}{4} = \frac{8!}{3!5!} \cdot \frac{5!}{2!3!} \cdot 3! \cdot \frac{5!}{4!1!} = \frac{8!}{3!2!} \cdot \frac{5!}{4!}$$

Why is your solution wrong?

The negation of all the vowels occur together is at least two of the vowels are separated. Therefore, if you wanted to count the complementary event, you would have to subtract all the ways in which at least two of the vowels are separated from the number of distinguishable arrangements of the letters of the word $INDEPENDENCE$. There are $$\binom{12}{4}\binom{8}{3}\binom{5}{2}3! = \frac{12!}{4!3!2!} = \frac{12!}{4!3!2!1!1!1!}$$ distinguishable arrangements of the word INDEPENDENCE. If at least two of the vowels are separated, you would have to take into account the following cases: \begin{align*} 5 & = 4 + 1\\ & = 3 + 2\\ & = 3 + 1 + 1\\ & = 2 + 2 + 1\\ & = 2 + 1 + 1 + 1\\ & = 1 + 1 + 1 + 1 + 1 \end{align*} where the numbers on the right-hand side of the equation represent the number of vowels in each isolated block of vowels. You would also have to take into account which block contained the I. Therefore, a direct count is much easier.

Answer 2

There are 5 vowels and 4 of them are same( 4 E's). So these 5 vowels can be arranged in $\frac{5!}{4!}$ ways

Answer 3

Since $IEEEE$ is always together we can 'group' them together as one (Let's call this group $V$).

Then in $V$ itself, there is ${5 \choose 1}= \frac{5!}{4!}$ ways. (This is where this fraction came from! We choose 1 place from the 5 to put the $I$ [and naturally we put $E$ for the other 4])

Now, The number of objects:

3 $N$

2 $D$

$P$

$C$

$V$ (with $\frac{5!}{4!}$ ways)

Hence the number of permutations = $\frac{8!}{3!2!} \times \frac{5!}{4!}$, where we divide $3!2!$ because $N$ was repeated 3 times, and $D$ was repeated 2 times.