Question

A five-digit number is formed by using the digits 1, 2, 3, 4, 5 with no repetition. The probability that the numbers 1 and 5 are always together, is

• A. \( \frac{2}{5} \)
• B. \( \frac{1}{5} \)
• C. \( \frac{3}{5} \)
• D. \( \frac{1}{4} \)

Hint:

When considering the probability of two specific numbers being together in an arrangement, treat them as a single unit (block) and find the total number of favorable outcomes.

Answer 1

The Correct Option is A

We need to determine the probability that in a five-digit number formed by using the digits 1, 2, 3, 4, 5 without repetition, the numbers 1 and 5 are always together. Let's solve this step by step:

1. Total number of arrangements: There are 5 digits available, so the total number of different five-digit numbers is given by the permutation of 5 digits:

   \(5! = 120\)

2. Considering 1 and 5 as a single entity: If we treat the numbers 1 and 5 as a single entity or "block", we effectively have 4 entities to arrange: the "1-5 block", 2, 3, and 4. Therefore, the number of ways to arrange these 4 entities is:

   \(4! = 24\)

3. Arrangement within the block: Within the "1-5 block", the numbers 1 and 5 can be arranged among themselves in 2 ways (15 or 51):

   \(2! = 2\)

4. Total favorable outcomes: The total number of arrangements where 1 and 5 are always together is:

   \(4! \times 2! = 24 \times 2 = 48\)

5. Probability: The probability that 1 and 5 are always together is the ratio of favorable outcomes to the total number of outcomes:

   \(\frac{48}{120} = \frac{2}{5}\)

Therefore, the probability that the numbers 1 and 5 are always together is \( \frac{2}{5} \).

Answer 2

The total number of ways to arrange the five digits \( 1, 2, 3, 4, 5 \) without repetition is: \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \] Now, we want to find the probability that the digits 1 and 5 are always together. To solve this, we can treat the pair \( 1, 5 \) as a single "block," so we have the following elements to arrange: \[ (1,5), 2, 3, 4 \] Thus, we have 4 elements to arrange, which can be done in \( 4! \) ways: \[ 4! = 4 \times 3 \times 2 \times 1 = 24 \] Within the "block" \( (1,5) \), the digits 1 and 5 can be arranged in \( 2! \) ways: \[ 2! = 2 \times 1 = 2 \] So, the total number of favorable outcomes is \( 4! \times 2! = 24 \times 2 = 48 \). The probability is given by the ratio of favorable outcomes to the total number of outcomes: \[ \text{Probability} = \frac{48}{120} = \frac{2}{5} \] Thus, the probability that the numbers 1 and 5 are always together is \( \frac{2}{5} \).