JEE Advanced Mathematics question
The number of numbers of 9 different nonzero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle is
Answer: (B)
Solution
Nine different nonzero digits means all of 1, 2, …, 9 are used. The middle digit must have exactly four digits smaller and four larger than it, so it must be 5.
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