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:

To handle "never together" problems for a subset of items, use the gap method: arrange the other items first to create slots (gaps), and then place the restricted items into those slots using permutations. The number of gaps is always one more than the number of items you arrange first.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Column A requires finding the number of permutations with a constraint (vowels not together), a common type of arrangement problem. Column B is a straightforward permutation of distinct letters.
Step 2: Key Formula or Approach:
For Column A, we use the "gap method". First, arrange the items that can be together (consonants), creating gaps. Then, place the restricted items (vowels) into these gaps. For Column B, the number of arrangements of \(n\) distinct items is \(n!\).
Step 3: Detailed Explanation:
For Column A:
The word is PROMISE. Total letters = 7. Vowels = O, I, E (3 vowels). Consonants = P, R, M, S (4 consonants). The condition is that no two vowels come together.
Step i: Arrange the consonants. The 4 distinct consonants (P, R, M, S) can be arranged in \(4!\) ways. \[ 4! = 4 \times 3 \times 2 \times 1 = 24 \] Step ii: Place the vowels in the gaps created by the consonants. The arrangement of consonants creates 5 possible gaps for the vowels (one before the first consonant, three between them, and one after the last). \[ \_ C \_ C \_ C \_ C \_ \] We need to place the 3 distinct vowels (O, I, E) in these 5 gaps. The number of ways to do this is a permutation of 5 items taken 3 at a time, \(P(5, 3)\). \[ P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = 5 \times 4 \times 3 = 60 \] Step iii: The total number of arrangements is the product of the ways from Step i and Step ii. \[ \text{Total arrangements} = (\text{Ways to arrange consonants}) \times (\text{Ways to place vowels}) \] \[ \text{Total arrangements} = 24 \times 60 = 1440 \] For Column B:
The given letters are I, N, U, R, E, V, L, A, S. There are 9 letters in total. All the letters are distinct. The number of words that can be formed is the number of permutations of these 9 letters. \[ \text{Total words} = 9! \] \[ 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362,880 \] Step 4: Final Answer:
Comparing the two quantities:
Quantity A = 1440
Quantity B = 362,880
Quantity B is significantly greater than Quantity A.