How many numbers exceeding 999 and not exceeding 5000 can be formed

Question

How many numbers exceeding 999 and not exceeding 5000 can be formed, such that each digit is any of the first six whole numbers?

A. 600
B. 625
C. 820
D. 864
E. 865

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IanStewart · Answer

The number must be less than 5000, because our digits cannot be zero, so we have 4 choices for the thousands digit, and 6 choices for each of the remaining three digits, for 4*6*6*6 possible numbers in total. The units digit of that product will be 4, so D must be right (or you could multiply it out).

edit: I was assuming the question meant 'natural numbers' instead of 'whole numbers' -- you will never need to know what either of those terms mean for the GMAT, and I've confused the two my whole life.

The 'whole numbers' include zero, so we actually do need to include '5000' here as well, and the answer should be E.

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The 'whole numbers' include zero, so we actually do need to include '5000' here as well, and the answer should be E.

kevinluume · Answer

Answer is E.

Numbers are exceeding 999 and  not exceeding  5000. This would mean that the range of numbers would be from 1000-5000. Digits are from 1-6.

Since there are a 4 place holders we can fill.

1-4,1-6,1-6,1-6,1-6 or 4x6x6x6=864
we would need to add one to account for 5000. So answer is 864+1=865

IanStewart · Answer

Your answer is right, and mine was not because I was assuming the digits were natural numbers rather than whole numbers, so I excluded '5000' from my count, though the digits here range from 0-5 and not from 1-6 (I was assuming it was 1-6 also!). But a real GMAT question would never be worded this way anyway.

How many numbers exceeding 999 and not exceeding 5000 can be formed

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How many numbers exceeding 999 and not exceeding 5000 can be formed

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