Understanding Permutations with Constraints
When we are asked to arrange the letters of a word such that specific items must be together (like all vowels), we use the "String Method" or "Bundle Method".
Step 1: Analyze the word "PRECARIOUS"
- Total letters: 10
- Letters are: P, R, E, C, A, R, I, O, U, S
- Identifying vowels: E, A, I, O, U (5 vowels)
- Identifying consonants: P, R, C, R, S (5 consonants)
Step 2: Bundle the vowels
Since the vowels must always be together, we treat the group {E, A, I, O, U} as a single unit or object.
Now, our set of items to arrange is: {P, R, C, R, S, (EAIOU)}
This gives us 6 items to arrange.
Step 3: Arrange the items
Among the 6 items, notice there are two 'R's. The number of ways to arrange these 6 items is given by the formula for permutations with repetitions: $\frac{n!}{n_1!n_2!...}$
Ways to arrange the 6 items = $\frac{6!}{2!} = \frac{720}{2} = 360$.
Step 4: Arrange the vowels inside the bundle
Inside the bundle {E, A, I, O, U}, there are 5 distinct vowels. They can be arranged among themselves in $5!$ ways:
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
Step 5: Final Calculation
By the Fundamental Counting Principle, the total number of arrangements is the product of the two steps:
Total ways = $360 \times 120 = 43,200$.