Solving Word Arrangement Problems: OPTICAL Example

Let's figure out how many different ways we can arrange the letters of the word 'OPTICAL' so that the vowels always stay together. This is a classic permutation problem with a constraint.

Understanding the Word and its Letters

The word is 'OPTICAL'. Let's break it down:

• Total number of letters: 7
• The letters are: O, P, T, I, C, A, L

Now, let's identify the vowels and consonants in 'OPTICAL':

• Vowels: O, I, A (There are 3 vowels)
• Consonants: P, T, C, L (There are 4 consonants)

Applying the Constraint: Vowels Together

The problem requires that the vowels (O, I, A) must always stay together. To handle this constraint, we can treat the group of vowels as a single unit or a single "block".

So, instead of having 7 individual letters to arrange, we now have the following items to arrange:

• The block of vowels (OIA)
• The 4 individual consonants (P, T, C, L)

This means we are effectively arranging 1 (vowel block) + 4 (consonants) = 5 items.

Arranging the Items (Vowel Block and Consonants)

We have 5 distinct items to arrange. The number of ways to arrange \(n\) distinct items is given by \(n!\) (n factorial).

Here, we are arranging 5 items, so the number of arrangements is \(5!\).

Calculation:

\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

So, there are 120 ways to arrange the vowel block and the consonants.

Arranging the Vowels within the Block

The vowels O, I, and A are within their block, but they can arrange themselves in different ways inside this block. Since there are 3 distinct vowels (O, I, A) in the block, the number of ways they can arrange themselves is \(3!\).

Calculation:

\(3! = 3 \times 2 \times 1 = 6\)

So, the vowels can be arranged in 6 different ways within their block (e.g., OIA, OAI, IOA, IAO, AOI, AIO).

Calculating the Total Number of Arrangements

To find the total number of ways to arrange the letters of 'OPTICAL' such that the vowels always come together, we multiply the number of ways to arrange the items (vowel block + consonants) by the number of ways to arrange the vowels within their block.

Total arrangements = (Arrangements of 5 items) \(\times\) (Arrangements of 3 vowels within the block)

Total arrangements = \(5! \times 3!\)

Total arrangements = \(120 \times 6 = 720\)

Summary of the Solution

Here’s a quick look at the steps involved:

The total number of different ways the letters of the word 'OPTICAL' can be arranged so that the vowels always come together is 720.

Revision Table: Word Arrangement with Constraints

Additional Information: Permutations and Combinations

Understanding permutations and combinations is key to solving counting problems like this one.

• Permutations: This is about arrangement where the order matters. If you have \(n\) distinct objects and want to arrange \(r\) of them, the number of permutations is \(P(n, r) = \frac{n!}{(n-r)!}\). When arranging all \(n\) objects, it's \(P(n, n) = n!\). Our problem is a permutation problem because changing the order of letters (like OPTICAL vs. TOPICAL) results in a different arrangement.
• Combinations: This is about selection where the order does NOT matter. If you have \(n\) distinct objects and want to choose \(r\) of them, the number of combinations is \(C(n, r) = \frac{n!}{r!(n-r)!}\). Our problem is not a combination problem because the order of letters in a word is important.

When solving problems involving arranging letters of a word with constraints (like vowels together, consonants together, specific letters at ends, etc.), the general approach is to first handle the constraint (often by grouping) and then calculate the permutations of the resulting units and the permutations within the grouped units.