Question

Steel Express stops at six stations between Howrah and Jamshedpur. Five passengers board at Howrah. Each passenger can get down at any station till Jamshedpur. The probability that all five persons will get down at different stations is:

• A. \(\dfrac{^6P_5}{6^5}\)
• B. \(\dfrac{^6C_5}{6^5}\)
• C. \(\dfrac{^7P_5}{7^5}\)
• D. \(\dfrac{^7C_5}{7^5}\)

Hint:

Jamshedpur itself counts as a valid stopping station, so there are 7 stations in total for each passenger, not 6.

Answer 1

The Correct Option is C

Step 1: Count the total stopping points.
Steel Express stops at six stations between Howrah and Jamshedpur, and it also stops at Jamshedpur itself. So each passenger has 6 + 1 = 7 stations where they can get down.

Step 2: Count the total number of ways all five can get down.
Each of the 5 passengers picks one of these 7 stations independently, so the total number of ways to assign stations is \(7^5\).

Step 3: Count the favourable ways (all different stations).
For all five passengers to get down at five different stations, we need an ordered arrangement of 5 stations chosen from the 7 available, one station per passenger. This is a permutation, so the count is \(^7P_5\).

Step 4: Write the probability.
Probability = favourable outcomes divided by total outcomes = \(\dfrac{^7P_5}{7^5}\). This also shows why the other options fail: the options with 6 in the base ignore Jamshedpur as a valid stop, and the options with \(^7C_5\) or \(^6C_5\) count unordered groups of stations instead of assigning a specific station to each passenger.

Final Answer:
The probability is \[ \boxed{\dfrac{^7P_5}{7^5}} \]