Question:

How many numbers greater than 50000 can be formed by using digits 2, 5, 5, 6, 7?

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When digits are repeated, use \(\frac{n!}{r!}\) and eliminate cases violating the given condition.
Updated On: Apr 15, 2025
  • 60
  • 48
  • 52
  • 42
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The Correct Option is B

Solution and Explanation


We are to form 5-digit numbers using the digits 2, 5, 5, 6, 7. Since '5' is repeated, total permutations:
\[ \frac{5!}{2!} = 60 \]
Now, to count numbers > 50000, the first digit must be either 5, 6, or 7. We’ll subtract the cases where the first digit is less than 5. Only digit > 5 is 2.
Case 1: First digit = 2
Remaining digits = 5, 5, 6, 7 → number of arrangements = \(\frac{4!}{2!} = 12\)
Required count: Total - Invalid = \(60 - 12 = 48\)
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