Question: How many three-digit integers greater than 710 are there such that all their digits are different?

1. 198
2. 202
3. 207
4. 209
5. 212

Answer:

Approach Solution 1:

First find how many integers between 200 and 999 are such that all their digits are different.

We have: (3 options for the first digit) * (9 options for the second digit) * (8 options for the third digit) = 216 numbers

Among these 216 numbers, 9 (701, 702, 703, 704, 705, 706, 708, 709, 710) are not bigger than 710. The answer to the question is therefore 216 – 9 = 207

Correct option: C

Approach Solution 2:

We have 3 options (7, 8, 9) for the first digit.

1) 188 = 64 ----- 7 is the first digit here
2) 198 = 72 ------ 8 is the first digit here
3) 198 = 72 ------- 9 is the first digit here

So total is 64+72+72 = 208.
However, in the first case we have to remove one number to account for 710. So answer is 207.

Correct option: C

“How many three-digit integers greater than 710 are there such that all their digits are different?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

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