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Earlier discussion
1 older comment kept for context.
Victor Titus
Note that every 4- digit and 5- digit numbers formed is greater than 150. Therefore, \(^{5}P_{5} + ^{5}P_{4} = \frac{5!}{(5 - 5)!} + \frac{5!}{(5 - 4)!} = 120 + 120 = 240\) numbers are greater than 150 already.
Among the 3- digit numbers, the numbers 123, 124, 125, 132, 134, 135, 142, 143 and 145 are removed from the ones that meet the criterion so we have:
\(^{5}P_{3} - 9 = \frac{5!}{(5 - 3)!} = 60 - 9 = 51 \implies 240 + 51 = 291\) numbers are greater than 150.