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RenB
Joined: 13 Jul 2022
Last visit: 14 Nov 2025
Posts: 391
Given Kudos: 303
Location: India
Concentration: Finance, Nonprofit
Schools: Stanford '26 Wharton '26 (D) Kellogg '26 (II) Booth '26 (D) Yale '26 (S) CBS '26 (II) Darden '26 (S) HBS '26 (D) Ross '26 (II) LBS '26 (D) Tuck '26 (II)
GMAT Focus 1: 715 Q90 V84 DI82 
GPA: 3.74
WE:Corporate Finance (Consulting)
Schools: Stanford '26 Wharton '26 (D) Kellogg '26 (II) Booth '26 (D) Yale '26 (S) CBS '26 (II) Darden '26 (S) HBS '26 (D) Ross '26 (II) LBS '26 (D) Tuck '26 (II)
GMAT Focus 1: 715 Q90 V84 DI82
Posts: 391
20 May 2023, 21:20
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
47% (02:24) correct
53%
(02:37)
wrong
based on 174
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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
A. 162
B. 215
C. 216
D. 240
E. 324
21 May 2023, 23:06
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RenB
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
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
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
