Question: How many three-digit integers greater than 710 are there such that all their digits are different?
- 198
- 202
- 207
- 209
- 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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