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How many different 3-digit . . .

Expert replies
by Vincen » Wed Sep 27, 2017 6:42 pm
How many different 3-digit numbers are greater than 299 and do not contain the digits 1, 6, or 8?

(A) 222
(B) 245
(C) 291
(D) 315
(E) 343

The OA is B.

I tried to make a list but it has so much numbers and I got tired. Is there an easier way to solve this PS question?
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Source: — Problem Solving |

by [email protected] » Wed Sep 27, 2017 7:41 pm
Hi Vincen,

This prompt is essentially just a 'permutation' question, but it does have some minor 'quirks' to it that you have to be careful about to get the correct answer.

We're asked for the total number of 3-digit numbers that:
1) Can only use the digits 0, 2, 3, 4, 5, 7, 9
2) Must be GREATER than 299.

These restrictions will obviously limit the number of possibilities, but there's one other restriction that we have to remember: a 3-digit number CANNOT begin with a 0 (e.g. 045 is "45", which is a 2-digit number).

The first digit can be: 3, 4, 5, 7, 9 (5 options)
The second digit can be: 0, 2, 3, 4, 5, 7, 9 (7 options)
The third digit can be: 0, 2, 3, 4, 5, 7, 9 (7 options)

(5)(7)(7) = 245 total options

Final Answer: B

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by Vincen » Fri Sep 29, 2017 7:19 pm
Thanks. I really appreciate it.
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