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Q.
How many numbers greater than $40000$ can be formed from the digits $2, 4, 5, 5, 7$ ?
Rajasthan PETRajasthan PET 2004
Solution:
Number of permutations formed by given numbers $ =\frac{5!}{2!} $ $ =5\times 4\times 3=60 $ Since, in these permutations numbers start with 2 are also included. Which will be less than 40000. If 2 be the starting digit, then the number of permutations formed by remaining four numbers $ =\frac{4!}{2!}=4\times 3=12 $
Hence, required permutations $ =60-12=48 $