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Akshay1298
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Of the three-digit positive integers that have no digits 12 Jun 2025, 05:55
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I'm not sure how correct was my solution but I have a question,

For sequence XXY,

the first digit has 9 possibilities
the second digit has 9 possibilities
the third digit has 8 possibilities

For one sequence = 9*9*8 possibilities;

The same will follow for the sequences XYX and YXX.

Question: Firstly, I multiplied 3*9*9*8 and it's not a part of the answer choices but by dividing (9*9*8)/3, I get the answer 216. How it works and why the multiplication didn't work?
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Of the three-digit positive integers that have no digits 13 Jun 2025, 00:38
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Akshay1298

Bunuel
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?

A. 24
B. 36
C. 72
D. 144
E. 216

Consider a three-digit integer in the form XXY.

  • Choices for digit X: 9 (since it can be any digit from 1 to 9).
  • Choices for digit Y: 8 (any digit from 1 to 9, excluding X).
  • Permutations of the digits in XXY: 3!/2! (since the two X's are identical).

Total combinations: 9*8*3!/2! = 216.

Answer: E.

Hope this helps.
I'm not sure how correct was my solution but I have a question,

For sequence XXY,

the first digit has 9 possibilities
the second digit has 9 possibilities
the third digit has 8 possibilities

For one sequence = 9*9*8 possibilities;

The same will follow for the sequences XYX and YXX.

Question: Firstly, I multiplied 3*9*9*8 and it's not a part of the answer choices but by dividing (9*9*8)/3, I get the answer 216. How it works and why the multiplication didn't work?
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Your method treats both Xs as separate choices, but they’re supposed to be the same digit. You only choose X once (9 options), not twice. So it should be 9 * 8, not 9 * 9 * 8. Then you account for different permutations of XXY with 3!/2!. Please check the discussion on previous page.

Hope it helps.
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Of the three-digit positive integers that have no digits 28 Sep 2025, 10:10
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Hi,

Can you please tell whats wrong with my approach:

XXY = 9x9x8/3!

Thanks

Bunuel
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?

A. 24
B. 36
C. 72
D. 144
E. 216

Consider a three-digit integer in the form XXY.

  • Choices for digit X: 9 (since it can be any digit from 1 to 9).
  • Choices for digit Y: 8 (any digit from 1 to 9, excluding X).
  • Permutations of the digits in XXY: 3!/2! (since the two X's are identical).

Total combinations: 9*8*3!/2! = 216.

Answer: E.

Hope this helps.
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Bunuel
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Of the three-digit positive integers that have no digits 28 Sep 2025, 11:01
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MegB07
Hi,

Can you please tell whats wrong with my approach:

XXY = 9x9x8/3!

Thanks


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That approach doesn’t make sense. The second 9 is wrong because you don’t choose X twice, once is enough. And dividing by 3! is also wrong, we have two identical X’s and one Y, so we multiply by 3!/2! = 3, which gives the three arrangements: XXY, XYX, and YXX.
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Of the three-digit positive integers that have no digits 14 Oct 2025, 03:34
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Let's break down what we're looking for:

We need three-digit numbers where:
  • All digits are 1-9 (no zeros)
  • Exactly two digits are the same
  • One digit is different from the other two

Think of examples like 112, 343, 787, or 565—each has exactly two matching digits and one different digit.

Here's the key insight you need to see:

In a three-digit number ABC (where A = hundreds, B = tens, C = units), the repeated digit can appear in exactly three different patterns:

  1. Pattern 1: A = B ≠ C (like 112, 334, 557)
  2. Pattern 2: A = C ≠ B (like 121, 343, 575)
  3. Pattern 3: B = C ≠ A (like 211, 433, 755)

These are the only ways to arrange exactly two matching digits in a three-digit number.

Now let's count Pattern 1 systematically:

For numbers where A = B ≠ C:
  • Choose the repeated digit (A = B): You can pick any digit from 1 to 9 → \(9\) choices
  • Choose the different digit (C): You can pick any digit from 1 to 9, except it must be different from the repeated digit → \(8\) choices

Total for Pattern 1: \(9 \times 8 = 72\) numbers

Notice that the same logic applies to Pattern 2 and Pattern 3—each pattern also gives us exactly 72 possible numbers.

Final calculation:

Since these three patterns are completely separate (no number appears in more than one pattern), we add them:

\(72 + 72 + 72 = 216\)

Answer: E (216)

Want to master the systematic framework for all counting problems like this? You can check out the complete step-by-step solution on Neuron by e-GMAT to understand the underlying pattern recognition strategy that works across similar combinatorics questions. You can also explore detailed solutions for other GMAT official questions on Neuron with comprehensive analytics to identify and strengthen your weak areas.

Hope this helps! 🎯
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Of the three-digit positive integers that have no digits 14 Aug 2026, 08:52
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I vouch for the approach mentioned by Bunuel but i took a slightly lengthy route which still could be helpful:

3-digit positive integers with no zero: each digit can be from 1-9 hence permutation is: 9*9*9 = 729
those with all three digits same: 111, 222, ......999 which are 9 such digits
those with all three distinct digits: 9*8*7 = 504 such digits.

729-9-504 = 216
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