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Incorrect OG12 answer - diagnostic PS Q.11

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by stormier » Wed Dec 22, 2010 9:01 am
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


I think the correct answer should be 89, which is not an option. The OA says C
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by GMATGuruNY » Wed Dec 22, 2010 9:17 am
stormier wrote: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


I think the correct answer should be 89, which is not an option. The OA says C
Hundreds and tens digit the same, units digit different:
Number of choices for hundreds digit = 3 (we can use 7, 8, or 9)
Number of choices for tens digit = 1 (must be the same as the hundreds digit)
Number of choices for units digit = 9 (we can use any digit other the digit already used)
Multiplying the number of choices for each digit, we get 3*1*9 = 27 integers.

Hundreds and units digit the same, tens digit different:
Number of choices for hundreds digit = 3 (we can use 7, 8, or 9)
Number of choices for units digit = 1 (must be the same as the hundreds digit)
Number of choices for tens digit = 9 (we can use any digit other the digit already used)
Multiplying the number of choices for each digit, we get 3*1*9 = 27 integers.

Hundreds digit different, tens and units digit the same:
Number of choices for hundreds digit = 3 (we can use 7, 8, or 9)
Number of choices for tens digit = 9 (we can use anything other than the digit already used)
Number of choices for units digit = 1 (must be the same as the tens digit)
Multiplying the number of choices for each digit, we get 3*9*1 = 27 integers.
But these 27 integers include 700, which we cannot include, leaving us 27-1 = 26 integers.

Total number of possible integers = 27+27+26 = 80.

The correct answer is C.
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by Rahul@gurome » Wed Dec 22, 2010 9:19 am
stormier wrote: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
Required number = (Number of 3 digit integers greater than 700) - (Number of 3 digit integers greater than 700 with 3 same digit) - (Number of 3 digit integers greater than 700 with different digits)

Number of 3 digit integers greater than 700 = (999 - 700) = 299
Number of 3 digit integers greater than 700 with 3 same digit = 3 (Namely 777, 888 and 999)
Number of 3 digit integers greater than 700 with different digits = 3*9*8 = 216

Thus required number = (299 - 216 - 3) = 80

The correct answer is C.
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by stormier » Wed Dec 22, 2010 9:51 am
Gentlemen - thanks. I triple counted 777, 888 and 999 resulting in 9 extra counts. Solving it by permutation is much quicker.
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by OneTwoThreeFour » Sat Feb 05, 2011 7:50 pm
Rahul@gurome wrote:
stormier wrote: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
Required number = (Number of 3 digit integers greater than 700) - (Number of 3 digit integers greater than 700 with 3 same digit) - (Number of 3 digit integers greater than 700 with different digits)

Number of 3 digit integers greater than 700 = (999 - 700) = 299
Number of 3 digit integers greater than 700 with 3 same digit = 3 (Namely 777, 888 and 999)
Number of 3 digit integers greater than 700 with different digits = 3*9*8 = 216

Thus required number = (299 - 216 - 3) = 80

The correct answer is C.
+1
I like the way you think. Never thought about it from this angle
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