Last visit was: 07 Sep 2026, 10:56 It is currently 07 Sep 2026, 10:56
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2 
705-805 (Hard)|   Combinations|                        
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 07 Sep 2026
Posts: 22,733
Own Kudos:
26,945
 [2]
Given Kudos: 302
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 22,733
Kudos: 26,945
 [2]
Of the three-digit integers greater than 700, how many have two digits 12 Apr 2018, 16:30
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36
Show more

We can solve this problem by analyzing the digits (0 to 9) that are being “doubled”.

If 0 is doubled, then the numbers can only be 800 and 900. So we have 2 numbers.

If 1 is doubled, then the numbers can only be 711, 811 and 911. So we have 3 numbers.

If 2, 3, 4, 5, or 6 is doubled, then we should have 3 numbers for each case since it’s analogous to 1 being doubled.

If 7 is doubled, then the numbers can only be 770, 771, 772, 773, 774, 775, 776, 778, 779; 707, 717, 727, 737, 747, 757, 767, 787, 797; 877, and 977. So we have a total of 20 numbers.

If 8 or 9 is doubled, then we should have 20 numbers for each case since it’s analogous to 7 being doubled.

Thus, altogether we have 2 + 6 x 3 + 3 x 20 = 2 + 18 + 60 = 80 such numbers.

Answer: C
User avatar
nick1816
User avatar
Retired Moderator
Joined: 19 Oct 2018
Last visit: 12 Mar 2026
Posts: 1,841
Own Kudos:
8,786
Given Kudos: 707
Location: India
Posts: 1,841
Kudos: 8,786
Of the three-digit integers greater than 700, how many have two digits 06 Oct 2019, 05:08
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We don't even have to solve this question. Just use the logic!!

Because of the symmetry, 3 digit integers that have two digits that are equal to each other and the remaining digit different from the other two from 700-799 is equal to that are from 800-899 or 900-999

Hence total 3 digit numbers greater than 700 have two digits that are equal to each other and the remaining digit different from the other two
= 3k-1 (as we have to exclude 700)

Only option C is in the form of 3k-1

Bunuel
Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36
Show more
User avatar
MonishBhawale
User avatar
Joined: 15 Jun 2021
Last visit: 29 Jul 2026
Posts: 32
Own Kudos:
36
Given Kudos: 270
Posts: 32
Kudos: 36
Of the three-digit integers greater than 700, how many have two digits 03 Jun 2023, 21:17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36
Show more

Arrangement method
9_ _ =(1*1*9 )*3!/2! = 27
same same different = 3!/2!
similarly for
8_ _ = 27
and 7 _ _ =27 but we want greater than 700 so minus 1 and the 27 will include 700 so minus 1 = 26

so
27+27+26 = 80
User avatar
DanTheGMATMan
User avatar
Joined: 02 Oct 2015
Last visit: 07 Sep 2026
Posts: 385
Own Kudos:
278
Given Kudos: 10
Expert
Expert reply
Posts: 385
Kudos: 278
Of the three-digit integers greater than 700, how many have two digits 29 May 2024, 20:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Interesting question, way easier to do total minus what can't happen:

­
User avatar
Purnank
User avatar
Joined: 05 Jan 2024
Last visit: 29 Apr 2026
Posts: 675
Own Kudos:
638
 [1]
Given Kudos: 167
Location: India
Concentration: General Management, Strategy
GMAT Focus 1: 635 Q88 V76 DI80
Products:
My Reviews
GMAT Focus 1: 635 Q88 V76 DI80
Posts: 675
Kudos: 638
 [1]
Of the three-digit integers greater than 700, how many have two digits 26 Jun 2024, 05:38
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
1st- with 7XX numbers
1. 7__ Here with same remaining two digits only possible combinations are (11,22,33,44,55,66,88,99) = 8 combinations
2. 7_7 Here 9 combinations are possible except 777.
3. 77_ Here as well 9 combinations are possible 
Total for 7xx 8+9+9 = 26
Simlary if we check for 8XX and 9XX we will get 27 for each as 800 and 900 is possible as it wasnt in 7xx case because of given condition in question "..greater than 700.."
Total possible cases = 26 + 27 + 27 = 80.
Answer C.
User avatar
lnyngayan
User avatar
Joined: 09 Mar 2023
Last visit: 13 Oct 2024
Posts: 27
Own Kudos:
2
Given Kudos: 239
Posts: 27
Kudos: 2
Of the three-digit integers greater than 700, how many have two digits 04 Sep 2024, 03:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
SOLUTION

Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. two digits are alike and third is different;
C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700);
A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9);
C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

Answer: C.
Show more
Could you explain more why A is 3x9x8?
What are the constraints for the second and third digits?

Many thanks!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 07 Sep 2026
Posts: 113,202
Own Kudos:
839,768
 [1]
Given Kudos: 111,359
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,202
Kudos: 839,768
 [1]
Of the three-digit integers greater than 700, how many have two digits 04 Sep 2024, 06:39
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
lnyngayan

Bunuel
SOLUTION

Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. two digits are alike and third is different;
C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700);
A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9);
C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

Answer: C.
Could you explain more why A is 3x9x8?
What are the constraints for the second and third digits?

Many thanks!
Show more
In case A, the digits are distinct. So, for the second digit, we cannot use the digit selected for the first one, and for the third digit, we cannot use the digits used for the first and second ones, hence 9 and 8 options, respectively.

Hope it's clear.
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 07 Sep 2026
Posts: 6,129
Own Kudos:
6,090
Given Kudos: 164
Location: India
Schools: HBS '28 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
My Reviews
Schools: HBS '28 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
Posts: 6,129
Kudos: 6,090
Of the three-digit integers greater than 700, how many have two digits 16 Feb 2025, 20:54
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

Let the number be of the form

Case 1: AAB in which A is same and B is different
A = {7,8,9} 3 options
B = 0-9 except A: 9 options
Total cases = 3*9 = 27

Case 2: ABA in which A is same and B is different
A = {7,8,9} 3 options
B = 0-9 except A: 9 options
Total cases = 3*9 = 27

Case 3: BAA in which A is same and B is different
B = {7,8,9} 3 options
A = 0-9 except A: 9 options
Total cases = 3*9 = 27

Total cases = 27+27+27=81

Since the three-digit integers are greater than 700, 700 is excluded from total cases.

Total cases = 81 - 1 = 80

IMO C
User avatar
sujoykrdatta
User avatar
Joined: 26 Jun 2014
Last visit: 23 Aug 2026
Posts: 598
Own Kudos:
1,252
Given Kudos: 14
Status:Mentor & Coach | GMAT Q51 | CAT 99.98
GMAT 1: 750 Q51 V39
Expert
Expert reply
GMAT 1: 750 Q51 V39
Posts: 598
Kudos: 1,252
Of the three-digit integers greater than 700, how many have two digits 29 Oct 2025, 16:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We need 2 digit same and one different
Possible cases: XXY | XYX | YXX

XXY: X has 3 options (7/8/9) and Y has 9 options = 27 cases
XYX: X has 3 options (7/8/9) and Y has 9 options = 27 cases
YXX: Y has 3 options (7/8/9) and X has 9 options = 27 cases (this includes 700, so one case has to be removed) i.e. 26 cases
Ans = 27 + 27 + 26 = 80 (B)


Bunuel
SOLUTION

Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. two digits are alike and third is different;
C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700);
A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9);
C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

Answer: C.
Show more
User avatar
sidchandan
User avatar
Joined: 24 Oct 2016
Last visit: 30 May 2026
Posts: 16
Own Kudos:
12
Given Kudos: 39
Posts: 16
Kudos: 12
Of the three-digit integers greater than 700, how many have two digits 01 Dec 2025, 10:17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel - S S D - We have to find out same, same, different - which can be arranged in 3! / 2! ways = 3 ways.

and for the blanks : Blank 1 is 7/8/9 - so 3 possibilities, Blank 2 is 1 possibility since the number is same, and Blank 3 is 9 possibilties since the number is position 1 and 2 cannot be used. So we have 3 X 1 X 9 = 27 ways X 3 ways of rearranging = 81 ways. But the correct answer is 80. Where am I incorrect in my understanding?

Bunuel
SOLUTION

Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. two digits are alike and third is different;
C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700);
A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9);
C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

Answer: C.
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 07 Sep 2026
Posts: 113,202
Own Kudos:
839,768
 [1]
Given Kudos: 111,359
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,202
Kudos: 839,768
 [1]
Of the three-digit integers greater than 700, how many have two digits 01 Dec 2025, 10:56
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sidchandan
Hi Bunuel - S S D - We have to find out same, same, different - which can be arranged in 3! / 2! ways = 3 ways.

and for the blanks : Blank 1 is 7/8/9 - so 3 possibilities, Blank 2 is 1 possibility since the number is same, and Blank 3 is 9 possibilties since the number is position 1 and 2 cannot be used. So we have 3 X 1 X 9 = 27 ways X 3 ways of rearranging = 81 ways. But the correct answer is 80. Where am I incorrect in my understanding?


Show more


The first digit has 3 options 7, 8, 9.

If the repeated digit is the same as the first digit, then the different digit has 9 options and we get 3 * 9. The second and third digits can switch, so that gives 3 * 9 * 2 = 54.

If the repeated digit is different from the first digit, then the repeated digit has 9 options. That gives 3 * 9 = 27. This would include 700, which is not allowed, so the total would be 26.

Therefore, 54 + 26 = 80.
User avatar
bhanu29
User avatar
GMAT Forum Director
Joined: 02 Oct 2024
Last visit: 07 Sep 2026
Posts: 826
Own Kudos:
500
Given Kudos: 746
Location: India
Schools: Anderson '27 Booth '27 Foster '27 Haas '27 Ross '27 Tepper '27 Stanford '27 CBS '27 NYU Stern Johnson (Cornell) McCombs '27 Kellogg '27 Sloan '27
GMAT Focus 1: 675 Q87 V85 DI79
GMAT Focus 2: 715 Q87 V84 DI86
GPA: 9.11
WE:Engineering (Technology)
Products:
My Reviews GMAT Club Tests Forum Quiz
Schools: Anderson '27 Booth '27 Foster '27 Haas '27 Ross '27 Tepper '27 Stanford '27 CBS '27 NYU Stern Johnson (Cornell) McCombs '27 Kellogg '27 Sloan '27
GMAT Focus 2: 715 Q87 V84 DI86
Posts: 826
Kudos: 500
Of the three-digit integers greater than 700, how many have two digits 25 Mar 2026, 11:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36
Show more

Observe the goal is to get to numbers of form 7XX 7X7 and 77X (excluding 700)

7XX -> 711,..766,788,799 (8 numbers)
7X7 -9 numbers (all except 7)
77X - 9 numbers (all except 7)

Total 26

For 8 _ _ and 9 _ _ you'll have 27 each because 800, 900 is included
So it's 80

Alternative method

total numbers 701 - 999 -> 299

all distinct -> 7 A B, 8 A B, 9 A B , 9*8*3 = 216
all same -> 3 numbers (777, 888, 999)

299 -216 - 3 = 80

Correct Answer C
User avatar
Adit_
User avatar
Joined: 04 Jun 2024
Last visit: 29 Jul 2026
Posts: 869
Own Kudos:
327
Given Kudos: 165
Posts: 869
Kudos: 327
Of the three-digit integers greater than 700, how many have two digits 27 Mar 2026, 07:48
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Number of ways = 9*2+9 for each of 700+, 800+ and 900+ = 3(27) = 81. And we need to subtract 1 for including 700.
Thus 81-1=80.

Answer: Option C
Bunuel
Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36
Show more
User avatar
nandini14
User avatar
Joined: 26 Jun 2025
Last visit: 07 Sep 2026
Posts: 58
Own Kudos:
3
Given Kudos: 7
Location: India
Concentration: Finance, International Business
Schools: Stanford '26 Kellogg LBS '27
GMAT Focus 1: 685 Q84 V84 DI84
GPA: 8
Schools: Stanford '26 Kellogg LBS '27
GMAT Focus 1: 685 Q84 V84 DI84
Posts: 58
Kudos: 3
Of the three-digit integers greater than 700, how many have two digits 17 Jun 2026, 09:54
Kudos
Add Kudos
Bookmarks
Bookmark this Post
It felt easier to use the forward method here-
Calculationg each for when its
1) 7 - 10+10+ 7
2) 8 - 10+10+ 7
3) 9 - 10+10+ 7
Total = 81 - will minus 1 to exclude 700
User avatar
commodinostrum
User avatar
Joined: 10 Dec 2025
Last visit: 04 Sep 2026
Posts: 5
Own Kudos:
4
Given Kudos: 18
Posts: 5
Kudos: 4
Of the three-digit integers greater than 700, how many have two digits 20 Aug 2026, 21:03
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The fastest way to solve this: due to symmetry, the same amount of number from 700-799, 800-899,900-999 fit the description, however 700 must be excluded, hence the correct answer+1 must be dividable by 3. C.80 is the only option: 81/3=27
   1   2 
Moderator:
Bunuel
Math Expert
113202 posts