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RenB
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Of the four-digit positive integers that are greater than 4000, how ma 20 May 2023, 21:20
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Of the four-digit positive integers that are greater than 4000, how many have three digits that are equal to each other and one digit that is not equal to the other three?

A. 162
B. 215
C. 216
D. 240
E. 324
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B
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Of the four-digit positive integers that are greater than 4000, how ma 21 May 2023, 23:06
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RenB
Of the four-digit positive integers that are greater than 4000, how many have three digits that are equal to each other and one digit that is not equal to the other three?

A. 162
B. 215
C. 216
D. 240
E. 324
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Case 1: The thousands digit repeats thrice

X _ _ _

The thousand's place can be selected in 6 ways ( one among 4, 5, 6, 7, 8, and 9).

As the digit in the thousand's place repeats thrice, we need another digit. This digit can be selected in 9 ways (all digits from 0 to 9 except the one selected in the thousand's place).

The last three positions of the number can be interchanged to form various arrangments, hence the total number of digits that can be formed with the thousand's digit repeating thrice =

\(6 * 9 * \frac{3!}{2!} = 162\)

Case 2: Thousands digit is not the repeating digit

X Y Y Y

The thousand's place can be selected in 6 ways ( one among 4, 5, 6, 7, 8, and 9).

The second digit can be selected in 9 ways (all digits from 0 to 9 except the one selected in the thousand's place).

The last three positions of the number can be interchanged to form various combinations, hence the total number of digits that can be formed with the thousand's digit repeating thrice =

\(6 * 9 * \frac{3!}{3!} = 54\)

This case also has the possibility of the number being 4000. However, as we need the integers that are above 4000, we need to subtract 1 from the obtained result.

The total number of digits that can be formed with the thousand's digit repeating thrice =

\(54 - 1 = 53\)

Total : 53 + 162 = 215

Option B

Similar Question to Practice :

https://gmatclub.com/forum/of-the-three ... 35188.html
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Of the four-digit positive integers that are greater than 4000, how ma 27 Apr 2025, 08:50
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Of the four-digit positive integers that are greater than 4000, how ma 11 Jun 2026, 15:00
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RenB
Of the four-digit positive integers that are greater than 4000, how many have three digits that are equal to each other and one digit that is not equal to the other three?

A. 162
B. 215
C. 216
D. 240
E. 324
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GMATNinja Video Solution:




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Of the four-digit positive integers that are greater than 4000, how ma 16 Aug 2023, 20:36
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2 cases. let 4 digit number be abcd i.e, 1000a+100b+10c+d where 0 <= b,c,d <= 9 and 1<= a <= 9

1. thousands digit is the equal digit.
now a has 6 possibilities - 4 to 9. among b,c,d - two are equal to a. the other digit has 9 possibilities (0 to 9 minus the digit that is a, since unequal). b,c,d has 2 equal elements and 1 unequal. arranging this will give 3 possibilities. so 6*(9*3) = 162

2. thousands digit is the unequal digit.
now a has 6 possibilities. b=c=d but not equal to a. so it has 9 possibilites. therefore 6*9 = 54. note that we have counted 4000 in this case. we need numbers greater than 4000. so minus 1. 53 possibilities.

thereofre total is 162+53 = 215 possibilities. answer is B
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Of the four-digit positive integers that are greater than 4000, how ma 30 Jun 2024, 23:58
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RenB
Of the four-digit positive integers that are greater than 4000, how many have three digits that are equal to each other and one digit that is not equal to the other three?

A. 162
B. 215
C. 216
D. 240
E. 324
Show more
­
Let's count from 4000 to 9999 including 4000, and then subtract 1 from the final result because we need to find numbers greater than 4000.

Out of the 4 digits, 3 digits are equal and other digit is not equal to these 3 digits, ruling out the case all 4 digits are equal.

Let's consider the first case: 4000 -> 4999,

For (4,0) combination -> we have the following 4 possibilities.

4000 | 4004 | 4044 | 4440 ---> 4 Possibilities

Similarly, for (4,1) -> (4,9) combinations; leaving out (4,4) combination. We have 4 Possibilities * 8 Other Combinations = 32 Possibilities.

Total for (4,0) , (4,1) , (4,2) , (4,3) , (4,5) , (4,6) ,(4,7) ,(4,8) ,(4,9) = 36 ways!

Now, for the total 4000 -> 9999,

We will have 6 cases - 4000 - 4999; 5000 - 5999; 6000-6999; 7000-7999; 8000-8999; 9000-9999 (6 cases)

We will have 36 Possibilities * 6 cases = 216 ways

Let's subtract 1, from 216 because we also considered 4000 as a possible number.

Final Answer = 216 - 1 = 215 ways!­
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Of the four-digit positive integers that are greater than 4000, how ma 01 Jul 2024, 00:23
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RenB
Of the four-digit positive integers that are greater than 4000, how many have three digits that are equal to each other and one digit that is not equal to the other three?
Case 1: 1st digit is different from other 3. 
Number of possibilities for first digit = 6    : {4,5,6,7,8,9}
Number of possibilities for other 3 digits = 9
Total cases = 6*9 = 54

Case 1: 1st digit is repeated 2 more times. 1 other digit is different from other 3. 
Number of possibilities for first digit = 6      : {4,5,6,7,8,9}
Number of possibilities for placement of these 3 digits = 3C2 = 3
Number of possibilities for chosing other digit = 9
Total cases = 6*3*9 = 162

We have to subtract the case of 4000 since it is not allowed. 

Of the four-digit positive integers that are greater than 4000, The number of integers that have three digits that are equal to each other and one digit that is not equal to the other three = 54 + 162 - 1 = 215

IMO B­
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Of the four-digit positive integers that are greater than 4000, how ma 02 Jul 2026, 14:10
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How I did it, kindly correct if wrong-

first digit can be chosen in 6 ways (4, 5, 6, 7, 8, 9), the second and third digit in 1 way because it has to be a repeat of the first digit.
the last digit can be chosen in 9 ways (excluding the digit chosen in the first three digits)

now the other two repeating digits which are similar to the first one can be given positions in 3! ways

the only thing one has to keep in mind is it is greater than 4000 hence it becomes 6*9*3*2 - 1 = 215
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Of the four-digit positive integers that are greater than 4000, how ma 03 Jul 2026, 02:13
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Hi ga821,

Your starting numbers are right: 6 ways for a thousands digit from {4,5,6,7,8,9} and 9 ways for the odd digit (any digit except the repeated one). The trouble is the arrangement factor and a missing case.

The overcount: don't arrange identical digits

When three of the digits are the same, you never rearrange them among themselves - swapping two identical digits gives you the same number. The only thing that actually moves is the single odd digit. So there's no 3! and no extra ×2 here; multiplying by 3*2 double-counts numbers that are literally identical.

Also, a quick arithmetic check: 6*9*3*2 = 324, so 324 - 1 = 323, not 215 - the formula as written doesn't land on the answer.

Why you actually need two separate cases

The ">4000" rule forces the thousands digit to be 4-9, and that constraint behaves differently depending on where the odd digit sits:

- Odd digit is NOT in the thousands place (thousands is part of the triple): 6 for the repeated digit × 9 for the odd digit × 3 positions for the odd digit among the last three slots = 162.
- Odd digit IS the thousands digit (last three are all the repeat): 6 for the thousands digit × 9 for the repeated digit × 1 (the other three slots are forced) = 54.

Total = 162 + 54 = 216, then subtract the single number 4000 - 215.

So the fix is: replace your ×3×2 with just "where does the one odd digit go," and add the case where that odd digit is the leading 4-9.

Lock in the identical-digit idea

How many distinct numbers use the digits 5, 5, 5, 7? It's not4! = 24. Just list them:

- 5557, 5575, 5755, 7555 - only 4.

You only ever decide where the odd digit goes - the identical 5's don't create new arrangements. That's exactly the overcount your extra factor introduced.

Answer: B

ga821
How I did it, kindly correct if wrong-

first digit can be chosen in 6 ways (4, 5, 6, 7, 8, 9), the second and third digit in 1 way because it has to be a repeat of the first digit.
the last digit can be chosen in 9 ways (excluding the digit chosen in the first three digits)

now the other two repeating digits which are similar to the first one can be given positions in 3! ways

the only thing one has to keep in mind is it is greater than 4000 hence it becomes 6*9*3*2 - 1 = 215
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Of the four-digit positive integers that are greater than 4000, how ma 03 Jul 2026, 13:22
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thank you so much, this really helped!
egmat
Hi ga821,

Your starting numbers are right: 6 ways for a thousands digit from {4,5,6,7,8,9} and 9 ways for the odd digit (any digit except the repeated one). The trouble is the arrangement factor and a missing case.

The overcount: don't arrange identical digits

When three of the digits are the same, you never rearrange them among themselves - swapping two identical digits gives you the same number. The only thing that actually moves is the single odd digit. So there's no 3! and no extra ×2 here; multiplying by 3*2 double-counts numbers that are literally identical.

Also, a quick arithmetic check: 6*9*3*2 = 324, so 324 - 1 = 323, not 215 - the formula as written doesn't land on the answer.

Why you actually need two separate cases

The ">4000" rule forces the thousands digit to be 4-9, and that constraint behaves differently depending on where the odd digit sits:

- Odd digit is NOT in the thousands place (thousands is part of the triple): 6 for the repeated digit × 9 for the odd digit × 3 positions for the odd digit among the last three slots = 162.
- Odd digit IS the thousands digit (last three are all the repeat): 6 for the thousands digit × 9 for the repeated digit × 1 (the other three slots are forced) = 54.

Total = 162 + 54 = 216, then subtract the single number 4000 - 215.

So the fix is: replace your ×3×2 with just "where does the one odd digit go," and add the case where that odd digit is the leading 4-9.

Lock in the identical-digit idea

How many distinct numbers use the digits 5, 5, 5, 7? It's not4! = 24. Just list them:

- 5557, 5575, 5755, 7555 - only 4.

You only ever decide where the odd digit goes - the identical 5's don't create new arrangements. That's exactly the overcount your extra factor introduced.

Answer: B


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