A) 12
B) 24
C) 36
D) 48
Correct Answer: D
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\[=60-12=48\]
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