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The given question is: Find the number of words that can be made by all of the letters in the word GEOMETRY so that no vowels are adjacent.

My approach: 5! × (6𝑃 3/2!) = 7200, using the logic of filling the blanks in _G_M_T_R_Y _

However, the given answer is 18,000. Where am I incorrect in my method (without using other methods such as the inclusive-exclusive principle)?

EDIT: Just for context, the working out provided by the answer key is 8!/2! - (6! × 3!/2!) = 18,000.

All help is appreciated.

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    $\begingroup$ Your answer is correct. There is an error in the answer key. $\endgroup$ Commented Aug 27, 2022 at 11:41
  • $\begingroup$ Well, I would have claimed that $y$ was a vowel in that word (what other vowel is in the last syllable?), but that change makes the number smaller, not larger. The official answer is far too large...there are only $\frac {8!}2=20,160$ unrestricted permutations after all. $\endgroup$
    – lulu
    Commented Aug 27, 2022 at 11:58
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    $\begingroup$ I, too, would treat it as a vowel."If the Y makes a hard Y sound (as in 'yes' or 'beyond'), Y is a consonant. If the Y makes a short I sound (as in 'myth' or 'gym'), Y is a vowel. If the Y makes a long I sound (as in 'my' or 'fly'), Y is a vowel. If the Y makes a long E sound (as in 'Germany' or 'hungry'), Y is a vowel." In any case, it is bad practice to use Y in such a question $\endgroup$ Commented Aug 27, 2022 at 12:41
  • $\begingroup$ See my comment following the answer of Nitin Uniyal. $\endgroup$ Commented Aug 27, 2022 at 13:12
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    $\begingroup$ @InanimateBeing: Apart from Y, W is also classified as a semi-vowel, i.e. the sound may be a vowel sound depending on its place in a word. readsters.com/wp-content/uploads/2010/12/… But it is my view that in combinatorics, we should only consider letters that are vowels as vowels, , viz. A,E,I,O,U. $\endgroup$ Commented Aug 27, 2022 at 19:25

2 Answers 2

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As has been pointed out in the comments, the letter Y is used as a vowel in the word GEOMETRY, which makes the word GEOMETRY an unfortunate choice for this exercise since the usual convention in combinatorics that A, E, I, O, U are vowels and the remaining letters of the alphabet are treated as consonants. Under that convention, your answer is correct. As Nitin Uniyal has indicated in his answer, $18,000$ is the correct answer to a different question, namely the number of arrangements of the word GEOMETRY in which the three vowels (other than Y) do not appear in three consecutive positions.

In what follows, we will adopt the convention that A, E, I, O, U are vowels and the remaining letters of the alphabet are consonants (even though that is not true in the word GEOMETRY).

Find the number of arrangements of all of the letters of the word GEOMETRY in which no two of the three vowels E, E, O are adjacent.

Method 1: We rephrase your method.

The five distinct consonants G, M, T, R, Y can be arranged in $5!$ ways. This creates six spaces in which we can place the vowels, four between successive consonants and two at the ends of the row. $$\square C_1 \square C_2 \square C_3 \square C_4 \square C_5 \square$$ To ensure that no two vowels are consecutive, we choose two of these six spaces for the two Es and one of the remaining four spaces for the O, which can be done in $\binom{6}{2}\binom{4}{1}$ ways. Hence, the number of arrangements of the letters of the word GEOMETRY in which no two of the three vowels E, E, O are adjacent is $$5!\binom{6}{2}\binom{4}{1} = 7200$$ as you found.

Method 2: We apply the Inclusion-Exclusion Principle.

If we initially ignore the restrictions, we have eight positions to fill with the letters E, E, G, M, O, R, T, Y. We can select two of the eight spaces for the Es, then arrange the remaining six distinct letters in the remaining six spaces in $$\binom{8}{2}6!$$ ways.

From these arrangements, we must subtract those arrangements in which at least one pair of vowels is adjacent.

A pair of vowels is adjacent: There are two possibilities, either the two Es are adjacent or an E and the O are adjacent.

The two Es are adjacent: We have seven objects to arrange: EE, G, M, O, R, T, Y. Since the seven objects are all distinct, they can be arranged in $7!$ ways.

An E and the O are adjacent: We have seven objects to arrange: E, G, M, R, T, Y, and a block containing an E and the O. Since the seven objects are all distinct, they can be arranged in $7!$ ways. The E and the O within the block can be arranged in $2!$ ways. Hence, there are $7!2!$ such arrangements.

However, if we subtract the arrangements with a pair of adjacent vowels from the total, we will have subtracted each arrangement with two pairs of adjacent vowels twice, once for each way we could have designated one of the pairs of adjacent vowels as the pair of adjacent vowels. We only want to subtract such arrangements once, so we must add them to the total.

Two pairs of adjacent vowels: For this to occur (under the stated convention), the three vowels E, E, O must appear in consecutive positions. We have six objects to arrange: G, M, T, R, Y, and a block containing two Es and the O. Since the six objects are all distinct, they can be arranged in $6!$ ways. The three vowels E, E, O can be arranged within the block in $3$ distinguishable ways. Hence, there are $3 \cdot 6!$ such arrangements.

Thus, by the Inclusion-Exclusion Principle, the number of arrangements of the letters of the word GEOMETRY in which no two of the vowels E, E, O are adjacent is $$\binom{8}{2}6! - 7! - 7!2! + 3 \cdot 6! = 7200$$

Method 3: We use complementary counting.

As we showed above, there are $$\binom{8}{2}6!$$ distinguishable permutations of the letters of the word GEOMETRY.

From these, we must subtract those arrangements in which at least two of the letters E, E, O are adjacent.

Exactly two of the three vowels E, E, O are adjacent: The five distinct consonants G, M, R, T, Y can be arranged in $5!$ ways. This creates six spaces in which to place the vowels, four between successive consonants and two at the ends of the row. $$\square C_1 \square C_2 \square C_3 \square C_4 \square C_5 \square$$ There are two cases to consider: the two Es are adjacent or an E and an O are adjacent.

The two Es are adjacent: There are six spaces in which to place the block containing the two Es. Since the O cannot be adjacent to an E, it must be placed in one of the other five spaces. Hence, there are $$5! \cdot 6 \cdot 5$$ such arrangements.

An E and the O are adjacent: There are six spaces in which to place the block containing an E and the O, $2!$ ways to arrange the E and O within that block, and five spaces in which to place the other E. Hence, there are $$5! \cdot 6 \cdot 2! \cdot 5$$ such arrangements.

All three vowels E, E, O are consecutive: We have six objects to arrange: G, M, R, T, Y, and a block containing the vowels E, E, O. As we showed above, there are $3 \cdot 6!$ such arrangements.

Thus, the number of arrangements of the letters of the word GEOMETRY in which no two of the three vowels E, E, O are adjacent is $$\binom{8}{2}6! - 5! \cdot 6 \cdot 5 - 5! \cdot 6 \cdot 2! \cdot 5 - 3 \cdot 6! = 7200$$

Find the number of the arrangements of all of the letters of the word GEOMETRY in which the three vowels E, E, O do not appear in three consecutive positions.

We subtract the number of arrangements in which the three vowels do appear in three consecutive positions from the total, which yields $$\binom{8}{2}6! - 3 \cdot 6! = 18,000$$

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Total words=$\frac{8!}{2!}=20160$
Treating the set of vowels $\{E,O,E\}$as a single entity, number of words in which the 3 vowels are always together=$6!\times \frac{3!}{2!}=2160$
Subtracting them to get $20160-2160=18000$ words in which the three vowels are never together.
EDIT-Number of words in which 2 distinct vowels $\{O,E\}$ are together=$7!\times 2!=10080$.
Number of words in which 2 $E's$ are together =$\frac{71}{2!}=2520$
Required answer=$18000-10080-2520=5400$

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    $\begingroup$ You are counting all the vowels don't come together, which is different from no vowels are adjacent, which is what the question asks. $\endgroup$ Commented Aug 27, 2022 at 12:45
  • $\begingroup$ @trueblueanil true, however Nitin's flawed answer still has value, if he edits it. His answer allows the OP (i.e. original poster) to explain the mistake made by the problem composer. $\endgroup$ Commented Aug 27, 2022 at 13:12
  • $\begingroup$ Your calculation for the case in which the two distinct vowels E, O are together is incorrect since you have not taken into account the position of the block containing the letters E and O. $\endgroup$ Commented Aug 27, 2022 at 23:02
  • $\begingroup$ @N.F.Taussig....I missed the second E actually. Now edited. Thanks for pointing out! $\endgroup$ Commented Aug 28, 2022 at 0:56
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    $\begingroup$ @qwerty Your answer is correct. This answer is incorrect since Nitin Uniyal needed to subtract those cases with exactly two vowels together rather than those cases with two vowels together (which allows for the possibility that all three vowels were together). $\endgroup$ Commented Aug 28, 2022 at 1:51

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