Question

• A. The quantity on the left is greater
• B. The quantity on the right is greater
• C. Both are equal
• D. The relationship cannot be determined without further information

Hint:

Recognize different types of permutation problems. If there are repeated items, you must divide by the factorial of the counts of each repeated item. If you are selecting and arranging a subset of distinct items, use the \(P(n, r)\) formula.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Column A asks for the number of permutations of a set of letters where one letter is repeated. Column B asks for the number of permutations of a specific length (4 letters) from a set of distinct letters.
Step 2: Key Formula or Approach:
For Column A, the number of permutations of \(n\) objects, where \(p_1\) objects are identical of one kind, \(p_2\) are identical of a second kind, ..., \(p_k\) are identical of a \(k\)-th kind is: \[ \frac{n!}{p_1! p_2! \dots p_k!} \] For Column B, the number of permutations of \(r\) objects taken from a set of \(n\) distinct objects is: \[ P(n, r) = \frac{n!}{(n-r)!} \] Step 3: Detailed Explanation:
For Column A:
The given letters are I, I, H, O, N, R. Total number of letters, \(n = 6\). The letter 'I' is repeated 2 times. The number of different words that can be formed is: \[ \frac{6!}{2!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{720}{2} = 360 \] So, Quantity A is 360.
For Column B:
The word is NUMBER. The letters are N, U, M, B, E, R. All 6 letters are distinct. We need to form 4-letter words, without repetition. This is an arrangement of 4 letters chosen from 6 distinct letters. Here, \(n=6\) and \(r=4\). The number of 4-letter words is \(P(6, 4)\). \[ P(6, 4) = \frac{6!}{(6-4)!} = \frac{6!}{2!} = \frac{720}{2} = 360 \] So, Quantity B is 360.
Step 4: Final Answer:
Comparing the two quantities:
Quantity A = 360
Quantity B = 360
The two quantities are equal.