Question:medium

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:

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Jamshedpur itself counts as a valid stopping station, so there are 7 stations in total for each passenger, not 6.
Updated On: Jul 10, 2026
  • \(\dfrac{^6P_5}{6^5}\)
  • \(\dfrac{^6C_5}{6^5}\)
  • \(\dfrac{^7P_5}{7^5}\)
  • \(\dfrac{^7C_5}{7^5}\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Work out how many stations are actually available.
Between Howrah and Jamshedpur there are 6 stations, and Jamshedpur is also a valid stop, so every passenger has 7 possible stations to choose from.

Step 2: Assign stations to passengers one at a time.
The first passenger can get down at any of the 7 stations. For the second passenger to pick a station different from the first, only 6 choices remain. The third passenger then has 5 choices, the fourth has 4, and the fifth has 3.

Step 3: Multiply out the favourable count.
$7 \times 6 \times 5 \times 4 \times 3 = 2520$, which is exactly $^7P_5$.

Step 4: Divide by the total number of ways.
Without any restriction, each of the 5 passengers can independently pick any of the 7 stations, giving $7^5$ total outcomes. So the probability that all five get down at different stations is $\dfrac{2520}{7^5} = \dfrac{^7P_5}{7^5}$.

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