Calculating Permutations of Words

This question asks us to compare the number of distinct permutations of the letters in two words: ‘PERMUTATIONS’ and ‘COMBINATIONS’. We need to find the number of permutations for each word and then determine the relationship between these numbers.

Permutations of 'PERMUTATIONS' (let this be x)

The word ‘PERMUTATIONS’ has 12 letters.

Let's list the letters and their frequencies:

• P: 1
• E: 1
• R: 1
• M: 1
• U: 1
• T: 2 (appears twice)
• A: 1
• I: 1
• O: 1
• N: 1
• S: 1

The total number of letters is \(n = 12\). The letter 'T' is repeated 2 times (\(n_1 = 2\)). All other letters appear only once.

The formula for the number of distinct permutations of \(n\) objects where some objects are identical is given by:

\( \text{Number of permutations} = \frac{n!}{n_1! n_2! \dots n_k!} \)

In this case, for the word ‘PERMUTATIONS’, the number of permutations, x, is:

\( x = \frac{12!}{2!} \)

Permutations of 'COMBINATIONS' (let this be y)

The word ‘COMBINATIONS’ also has 12 letters.

Let's list the letters and their frequencies:

• C: 1
• O: 2 (appears twice)
• M: 1
• B: 1
• I: 2 (appears twice)
• N: 2 (appears twice)
• A: 1
• T: 1
• S: 1

The total number of letters is \(n = 12\). The letter 'O' is repeated 2 times (\(n_1 = 2\)), 'I' is repeated 2 times (\(n_2 = 2\)), and 'N' is repeated 2 times (\(n_3 = 2\)). All other letters appear only once.

Using the same formula for distinct permutations, for the word ‘COMBINATIONS’, the number of permutations, y, is:

\( y = \frac{12!}{2! 2! 2!} \)

Comparing x and y

We have \( x = \frac{12!}{2!} \) and \( y = \frac{12!}{2! 2! 2!} \).

Let's express y in terms of x:

\( y = \frac{12!}{2! \times 2! \times 2!} \)

\( y = \frac{12!}{2!} \times \frac{1}{2! \times 2!} \)

Since \( x = \frac{12!}{2!} \), we can substitute x into the equation for y:

\( y = x \times \frac{1}{2! \times 2!} \)

Calculate the factorials in the denominator:

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

So, \( 2! \times 2! = 2 \times 2 = 4 \).

Substitute this back into the equation for y:

\( y = x \times \frac{1}{4} \)

\( y = \frac{x}{4} \)

To find the relationship between x and y, we can rearrange the equation:

\( 4y = x \)

or

\( x = 4y \)

Thus, the number of permutations of ‘PERMUTATIONS’ (x) is equal to 4 times the number of permutations of ‘COMBINATIONS’ (y).

Conclusion

Comparing our result \( x = 4y \) with the given options, we find that the correct relationship is \( x = 4y \).

Revision Table: Permutations Calculation

Additional Information: Permutations and Combinations

Permutations and combinations are fundamental concepts in combinatorics, which is a branch of mathematics concerned with counting, arranging, and choosing objects.

• Permutations: A permutation is an arrangement of objects in a specific order. The order matters in permutations. The number of permutations of \(n\) distinct objects is \(n!\). If there are repetitions, the formula adjusts as shown in the solution.
• Combinations: A combination is a selection of objects where the order does not matter. The number of combinations of selecting \(k\) objects from a set of \(n\) distinct objects is given by \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
• Understanding factorials is crucial for both concepts. A factorial of a non-negative integer \(n\), denoted by \(n!\), is the product of all positive integers less than or equal to \(n\). \(0! = 1\).

This problem focuses specifically on permutations of words with repeated letters, a common application of permutation principles.