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Of the three-digit integers greater than 800

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Of the three-digit integers greater than 800 [#permalink] New post   02 Aug 2017, 08:55
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Of the three-digit integers greater than 800, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 36
(B) 51
(C) 53
(D) 54
(E) 60
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C
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Of the three-digit integers greater than 800 [#permalink] New post   02 Aug 2017, 10:05
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With the given conditions, three types of numbers are possible

Type 1: AAB -- First and Second Digits are same, third digit is different
Type 2: ABA -- First and Third Digits are same, Second digit is different
Type 3: BAA -- Second and third Digits are same, First digit is different

Type 1: First and Second Digit can be 8 and 9, Third Digit can be 0,1,2,3,4,5,6,7,8/9. (If first and second digit is 8, then third digit cannot be 8 and if first and second digit is 9, then third digit cannot be 9).

Therefore total no. of combinations for Type 1 = 2 *9 = 18

Type 2: First and Third Digit can be 8 and 9, Second Digit can be 0,1,2,3,4,5,6,7,8/9. (If first and third digit is 8, then Second digit cannot be 8 and if first and third digit is 9, then second digit cannot be 9).

Therefore total no. of combinations for Type 2 = 2 *9 = 18

Type 3: If First Digit is 8, then second and third digit can be 1,2,3,4,5,6,7,9. If First Digit is 9, then second and third digit can be 0,1,2,3,4,5,6,7,8.

Therefore total no. of combinations for Type 3 = 8+9 = 17

Total number of combinations = 18 + 18 + 17 = 53

Answer is c
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Re: Of the three-digit integers greater than 800 [#permalink] New post   03 Aug 2017, 02:52
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Turkish wrote:
Of the three-digit integers greater than 800, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 36
(B) 51
(C) 53
(D) 54
(E) 60


Three digit numbers have only 3 patterns:
1) all digits are alike
2) only 2 digits are alike
3) all digits are unique

Here, we shall calculate Desired (2 above) = Total - Undesired (1+3 above)


First let's find total digits between 801 <--> 999
Total Digits = 999-801+1 = 199

Now, digits distinct between 801 and 999 --> 2X9X8 = 144
AND, all three alike --> only 2 = 888 and 999

So, our desired answer is = total - undesired
= 199 - 144 - 2 = 199 - 146 = 53

Hope this helps! :)

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Re: Of the three-digit integers greater than 800 [#permalink] New post   02 Aug 2017, 23:11
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Turkish wrote:
Of the three-digit integers greater than 800, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 36
(B) 51
(C) 53
(D) 54
(E) 60


This is a copy of the following OG question: https://gmatclub.com/forum/of-the-three ... 35188.html

Similar questions:
https://gmatclub.com/forum/of-the-three ... 27390.html
https://gmatclub.com/forum/how-many-odd ... 71394.html

Hope it helps.
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Re: Of the three-digit integers greater than 800 [#permalink] New post   03 Aug 2017, 03:15
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Turkish wrote:
Of the three-digit integers greater than 800, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 36
(B) 51
(C) 53
(D) 54
(E) 60


8xx
tens digit 8 = 9 numbers
unit digit 8 = 9 numbers
unit and tens digit same = 8 (cannot take 8 or 0)

9xx
tens digit 9 = 9 numbers
unit digit 9 = 9 numbers
unit and tens digit same = 9 (cannot take 9)

Total 9*5 + 8 = 53
C
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Re: Of the three-digit integers greater than 800 [#permalink] New post   10 Aug 2017, 11:12
Total number of 3 digit numbers = 199
Total number of 3 digit numbers with all digits same = 2
Total number of digits will all digits different greater than 800 = 9 * 8= 72
Total number of digits will all digits different greater than 900 = 9*8 =72
199 -(2+72*2)=53_
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Re: Of the three-digit integers greater than 800 [#permalink] New post   19 Jun 2021, 07:50
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Re: Of the three-digit integers greater than 800 [#permalink]
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