Consider a seven-letter word formed by mixing up the letters in the word COMBINE. How many ways can you do this if no vowel is isolated between two consonants?
[eg. EBMCION and MOIENCB are acceptable, but BEMCNIO is not]
The final answer is given to be $1872$.
We are being asked to find the numeric value of [Total number of seven-letter words] - [Number of words such that a vowel is isolated between two consonants]. The word "COMBINE" has $4$ consonants and $3$ vowels. To compute the number of constructed words with an isolated vowel between two consonants, I tried the following:
1st attempt:
We may generalize the case, representing consonants as $1$s and vowels as $0$s. Now, we are finding the total number of seven-digit binary strings with four $1$s and three $0$s, containing the substring $101$. There are $\frac{4!}{2!2!} = 6$ strings that can be formed from the remaining two $0$s and $1$s. We may insert the substring $101$ into any position in these 6 strings. We have five available positions, because there are four digits. Therefore, there are $6 \times 5=30$ seven-digit binary strings with four $1$s and three $0$s, containing the substring $101$.
Since each vowel and consonant is unique in the word COMBINE, there are $30 \times 4!\times3!=4320$ seven-letter words formed from COMBINE that contain an isolated vowel between two consonants, if we account for permutations of the vowels and consonants. However, $7!-4320=720 \neq 1872$, meaning my answer is incorrect.
2nd attempt:
Since there are three vowels, for there to be an isolated vowel between two consonants, either all vowels are separated, or two vowels may be grouped up together while the remaining vowel is isolated [there is no other case].
Case when all vowels are isolated:
First, we arrange the consonants, Then, we choose three out of five possible places to place
the vowels [.C.C.C.C., where C denotes a consonant and the periods represent possible places to put a vowel]. We finally account for the arrangement of the vowels. $$ P[4,4] \times C[5,3] \times P[3,3] = 1440.$$
Case when one vowel is isolated, while the remaining two are a pair:
First, we arrange the consonants. We then choose a vowel. Then, we choose a position out of three possible places to put it [C.C.C.C]. Then, we choose a position out of four possible places to
place the pair of the remaining vowels [.CVC.C.C. just as a possible example]. Finally, we account for the arrangement of the vowels in the pair. $$ P[4,4] \times C[3,1] \times C[3,1] \times C[4,1] \times P[2,2] = 1728.$$ $7!-[1440+1728] = 1872$, which confirms that this is the correct answer.
What is wrong with my first attempt? Where did I make my error?
Both of your answers are incorrect.
Method 1: We arrange the consonants, then place the vowels in the spaces between and at the ends of the row.
There are $4!$ ways of arranging the four distinct consonants B, G, L, R. For each such arrangement, there are five spaces in which we could place the vowels, three between successive consonants and two at the ends of the row. To separate the vowels, we must place the three vowels in three of these five spaces. Choose two of the five spaces for the As and one of the remaining three spaces for the E. Thus, the number of admissible arrangements is $$4!\binom{5}{2}\binom{3}{1} = 720$$
Method 2: We use the Inclusion-Exclusion Principle.
There are seven letters in total, including two As, one B, one E, one G, one L, and one R. There are $\binom{7}{2}$ ways to choose two of the seven positions in the arrangement for the As. The remaining five letters can be placed in the remaining five positions in $5!$ ways. Hence, there are a total of $$\binom{7}{2}5! = \frac{7!}{2!5!} \cdot 5! = \frac{7!}{2!}$$ ways of arranging the letters of the word ALGEBRA.
From these, we must subtract those arrangements in which one or more pairs of vowels are consecutive.
A pair of vowels is consecutive: Two As are consecutive or an A and E are consecutive.
Two As are consecutive: We have six objects to arrange. They are AA, B, E, G, L, R. Since they are distinct, they can be arranged in $6!$ ways.
An A and an E are consecutive: We have six objects to arrange. They are B, E, G, L, R, and a block containing an A and an E. The six objects are distinct, so they can be arranged in $6!$ ways. The A and E can be arranged within the block in $2!$ ways. Hence, there are $6!2!$ such arrangements.
If we subtract the number of arrangements in which a pair of vowels is consecutive from the total, we will have subtracted each arrangement in which two pairs of vowels are consecutive twice, once for each way we could have designated one of those pairs as the pair of consecutive vowels. Since we only want to subtract such arrangements once, we must add them to the total.
Two pairs of consecutive vowels: Since there are only three vowels, this can only occur if the three vowels are consecutive. We have five objects to arrange, B, G, L, R, and the block of three vowels. Since the objects are distinct, they can be arranged in $5!$ ways. The three vowels can be arranged in three ways: AAE, AEA, EAA. Hence, the number of such arrangements is $5! \cdot 3$.
By the Inclusion-Exclusion Principle, the number of admissible arrangements is $$\frac{7!}{2!} - 6! - 6!2! + 5! \cdot 3 = 720$$