kitchen table math, the sequel: help desk - combinations

Sunday, April 10, 2011

help desk - combinations

In the integer 3,589 the digits are all different and increase from left to right. How many integers between 4,000 and 5,000 have digits that are all different and that increase from left to right?
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4 comments:

rocky said...

4567 4568 4569

4578 4579

4589

4678 4679

4689

4789

There are ten.

How can I do this without using "brute force"?

April 10, 2011 at 11:49 AM
Anonymous said...

"between 4000 and 5000" means the first digit must be a 4 (since 5000 doesn't have the other properties.
The remaining digits must be 3 distinct digits from {5,6,7,8,9}, so there are "5 choose 3" of them. The order of the digits is fixed, so there's your answer: (5 choose 3) = 10.

April 10, 2011 at 1:49 PM
rocky said...

Thank you. I see it now. Once you choose the three digits, there is only one (ascending) order.

April 10, 2011 at 3:27 PM
Catherine Johnson said...

Once you choose the three digits, there is only one (ascending) order.

right - I didn't see that either

By happenstance, I had just reached the section in AofPS that has problems of this type, and I got all balled up reading this problem.

Didn't manage to even know what it was asking (or to notice that the second digit of every number would have to be 5, 6, 7, or 8).

April 10, 2011 at 7:07 PM

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