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705-805 (Hard)|   Combinations|                        
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Of the three-digit integers greater than 700, how many have two digits 12 Apr 2018, 16:30
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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
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Of the three-digit integers greater than 700, how many have two digits 06 Oct 2019, 05:08
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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
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Of the three-digit integers greater than 700, how many have two digits 03 Jun 2023, 21:17
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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
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Of the three-digit integers greater than 700, how many have two digits 29 May 2024, 20:51
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Interesting question, way easier to do total minus what can't happen:

­
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Of the three-digit integers greater than 700, how many have two digits 26 Jun 2024, 05:38
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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.
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Of the three-digit integers greater than 700, how many have two digits 04 Sep 2024, 03:21
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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.
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Could you explain more why A is 3x9x8?
What are the constraints for the second and third digits?

Many thanks!
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Of the three-digit integers greater than 700, how many have two digits 04 Sep 2024, 06:39
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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!
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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.
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Of the three-digit integers greater than 700, how many have two digits 16 Feb 2025, 20:54
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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
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Of the three-digit integers greater than 700, how many have two digits 29 Oct 2025, 16:34
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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.
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Of the three-digit integers greater than 700, how many have two digits 01 Dec 2025, 10:17
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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.
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Of the three-digit integers greater than 700, how many have two digits 01 Dec 2025, 10:56
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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?


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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.
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Of the three-digit integers greater than 700, how many have two digits 25 Mar 2026, 11:55
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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
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Of the three-digit integers greater than 700, how many have two digits 27 Mar 2026, 07:48
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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
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