How many different 3-digit numbers are greater than 299 and

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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

OA B

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BTGmoderatorDC wrote: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
Take the task of creating the 3-digit numbers and break it into stages.

Stage 1: Select the first digit (hundreds digit)
Since the first digit can be 3, 4, 5, 7 or 9, we can complete stage 1 in 5 ways

Stage 2: Select the second digit (tens digit)
Since the second digit can be 0, 2, 3, 4, 5, 7 or 9, we can complete stage 2 in 7 ways

Stage 3: Select the third digit (units digit)
Since the third digit can be 0, 2, 3, 4, 5, 7 or 9, we can complete stage 3 in 7 ways

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create a 3-digit number) in (5)(7)(7) ways (= 2445 ways)

Answer: B
--------------------------

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by Jeff@TargetTestPrep » Sat Aug 18, 2018 6:49 pm
BTGmoderatorDC wrote: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 first (or hundreds) digit has 5 choices (digits 3 to 9, excluding 6 and 8). Each of the second (or tens) and the third (or units) digits has 7 choices (digits 0 to 9, excluding 1, 6 and 8). Therefore, the number of 3-digit numbers greater than 299 that do not contain the digits 1, 6, or 8 is

5 x 7 x 7 = 245

Answer: B

Jeffrey Miller
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