Arrangements of TINKERER with restriction

Question

How many arrangements of TINKERER with two but not three consecutive vowels?

The total number of arrangements is $\displaystyle\frac{8!}{2!2!}=10,080$.

My approach is to use the complement and consider the number of arrangements with three consecutive vowels.

Step 1: Treat IEE as one group. Choose positions: $\binom{8}{3}=56$

Step 2: Arrange IEE: 3 ways

Step 3: Arrange the remaining letters: $\displaystyle\frac{5!}{2!}=60$ ways.

Step 4: Choose positions: $\binom{5}{5}=1$

I realized that this attempt did not work

Can someone please help!

Answer 1

How many arrangements of TINKERER with two but not three consecutive vowels?

My approach is to use the complement and consider the number of arrangements with three consecutive vowels.

Your approach: $n$(2, but not 3 consecutive vowels)=$n$(total)-$n$(3 consecutive vowels)-$n$(single vowels).

Total: $$\text{TINKERER} \\ n(\text{total})=\frac{8!}{2!2!}=10,080.$$

3 consecutive vowels: $$\boxed{\text{IEE}} \ T \ \square \ N \ \square \ K \ \square \ R \ \square \ R \ \square \\ n(3 \text{ consecutive})={6\choose 1}\cdot \frac{3!}{2!}\cdot \frac{5!}{2!}=1,080.$$ Interpretation: There are ${6\choose 1}$ positions to arrange the block of triple letters $\boxed{\text{IEE}}$. There are $\frac{3!}{2!}$ ways to arrange the three letters $I,E,E$ within the block. There are $\frac{5!}{2!}$ ways to arrange the rest letters $T,N,K,R,R$.

Single vowels: $$\boxed{\text{I}} \ T \ \boxed{\text{E}} \ N \ \boxed{\text{E}} \ K \ \square \ R \ \square \ R \ \square \\ n(\text{single vowels})=\frac{6!}{3!}\cdot \frac{1}{2!}\cdot \frac{5!}{2!}=3,600.$$ Interpretation: There are $\frac{6!}{3!}$ ways to arrange the three letters $I,E_1,E_2$ in $6$ positions and since the order of the two letters $E$ are not important, we multiply by $\frac{1}{2!}$. There are $\frac{5!}{2!}$ ways to arrange the rest letters $T,N,K,R,R$.

Hence: $n$(2, but not 3 consecutive vowels)$=10,080-1,080-3,600=5,400$.

Answer 2

You were on the right track. Take the case of complementary counting and $IEE$ is one block. The number of bad arrangements can be thought of when you just treat it as one "letter". However, your calculation of the arrangements is inccorect. First, you have to find the number of ways to arrange $IEE$ itself (inside the new "letter" we created). Clearly there are 3 ways, based on the position of $I$. Finally, we have to find the number of arrangements of $$TNKRR\underline{IEE}$$, which is $\frac{6!}{2!} = 360$, meaning there are $360*3 = 1080$ bad arrangements. Your argument could've worked if you fixed your first step to not $8$ choose $3$, $6 * 3$, where $6$ was the number of positions of $IEE$ (note there are not 8! You can only place the block $IEE$ in $6$ different places - try to convince yourself that this is true by attempting to find $8$ places to put it. $3$ is, like above, the number of ways to arrange $IEE$ in the new "letter" we have created.

Answer 3

How many distinguishable arrangements of the letters of the word $TINKERER$ have exactly two consecutive vowels?

We first arrange the consonants, then insert the vowels.

The five consonants are $K, N, R, R, T$. Choose two of the five positions for the $R$s. The remaining three distinct consonants can be arranged in the remaining three positions in $$\binom{5}{2}3! = \frac{5!}{2!3!} \cdot 3! = \frac{5!}{2!}$$ ways.

Arranging the consonants creates six spaces in which we can place the vowels. For example, $$\square K \square N \square R \square R \square T \square$$ To ensure that exactly two of the vowels are consecutive, we must place a block of two vowels in one of these six spaces and a single vowel in one of the other five spaces.

Since exactly two of the vowels are together, there are three possible ways to group the vowels. They are

1. the block $EE$ and the singleton $I$
2. the block $EI$ (in that order) and the singleton $E$
3. the block $IE$ (in that order) and the singleton $E$

Since there are $\binom{5}{2}3!$ ways to arrangements the consonants, $3$ ways to group the vowels, $6$ ways to place the block of two consecutive vowels, and $5$ ways to place the single vowel, the number of admissible arrangements is $$\binom{5}{2}3! \cdot 3 \cdot 6 \cdot 5 = 5400$$

Check: The number of ways to arrange the letters of the word $TINKERER$ is $$\binom{8}{2}\binom{6}{2}4! = 10080$$ as you found, since we must select two of the eight positions for the $E$s, two of the remaining six positions for the $R$s, and then arrange the remaining four distinct letters in the remaining four positions.

Notice that since there are three vowels in $TINKERER$, either no two vowels are consecutive, exactly two vowels are consecutive, or all three vowels are consecutive.

No two vowels are consecutive: We arrange the five consonants as above, which creates six spaces. To ensure that no two vowels are consecutive, we fill two of these six spaces with $E$s and one of the remaining four spaces with the $I$. Since there are $\binom{5}{2}3!$ ways of arranging the consonants, $\binom{6}{2}$ ways to place the $E$s, and $4$ ways to place the $I$, there are $$\binom{5}{2}3!\binom{6}{2}\binom{4}{1} = 3600$$ such arrangements.

All three vowels are consecutive: We arrange the five consonants as above, which creates six spaces in which we can place the block of three vowels. The three vowels can be arranged within the block in three ways, depending on whether $I$ is on the left, on the right, or in the middle. Since there are $\binom{5}{2}3!$ ways to arrange the consonants, $6$ ways to place the block of vowels, and $3$ ways to arrange the vowels within the block, there are $$\binom{5}{2}3!\binom{6}{1}\binom{3}{1} = 1080$$ such arrangements.

Total: Adding the number of arrangements in which no two vowels are consecutive, exactly two are consecutive, and all three are consecutive yields $$3600 + 5400 + 1080 = 10080$$ as we would expect.