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How many odd three-digit integers greater than 800 are there such that all their digits are different? (A) 40 (B) 56 (C) 72 (D) 81 (E) 104 Source: GMAT Club Tests - hardest GMAT questions Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker. For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part: For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3-digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)-5= 45-5= 40 Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4) - 4 = 36 - 4 = 32 possibilites. Add the 2 together, and you get 40+32=72. The answer! Hope that helps!!
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Hello, I know this answer has been recently discussed but that explanation confuses me even more. ____________________________________________________________________________________ How many odd three-digit integers greater than 800 are there such that all their digits are different? 72= Answer I understand the explanation given, the answer divided into two parts. What I don't understand is how you come up with 72 when you don't break the answer in two. The way I see it: There are 2 options for the first spot (8 and 9), 4 options for the last spot (1,3,5,7, without the 9 because of option in first spot), and 8 options for middle spot (because you start with ten options and minus 2 for the numbers taken in the other two spots)= 64. Why is it 2*9*4 instead? Why are there 9 options for the middle spot? Can someone please explain that? Thank you. 
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dczuchta wrote: Hello, I know this answer has been recently discussed but that explanation confuses me even more. ____________________________________________________________________________________ How many odd three-digit integers greater than 800 are there such that all their digits are different? 72= Answer I understand the explanation given, the answer divided into two parts. What I don't understand is how you come up with 72 when you don't break the answer in two. The way I see it: There are 2 options for the first spot (8 and 9), 4 options for the last spot (1,3,5,7, without the 9 because of option in first spot), and 8 options for middle spot (because you start with ten options and minus 2 for the numbers taken in the other two spots)= 64. Why is it 2*9*4 instead? Why are there 9 options for the middle spot? Can someone please explain that? Thank you. First place = 8 or 9 = 2 Second place = 0 - 7 and either 8 or 9 = 8 Third place = 1, 3, 5, 7 and 9 but there could be odd integer in first or second place. so; 1. if 8 in first place and 4 even integers in second place = 1x4x5 = 20 2. if 8 in first place and one of 5 odds in second place and one of remaining 4 odds in third place = 1x5x4 = 20 3. if 9 in first place and one of 5 evens in second place and one of 4 odds in third place = 1x5x4 = 20 4. if 9 in first place and one of 4 odds in second place and one of remaining 3 odds in third place = 1x4x3 = 12 total = 72.
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I used the following method during my test:
If 8 is first digit,
there are 5 ways to fill third digit (1,3,5,7 or 9) and 8 ways to fill 2nd digit: 5x8 = 40
If 9 is first digit,
there are 4 ways to fill 3rd digit (1,3,5 or 7) and 8 ways to fill 2nd digit: 4x8 = 32
40+32 = 72
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Hello,
Can you please help me understand as how there are 8 digits in the middle -
If first can be 8
then middle is 0,1,2,3,4,5,6,7,9 - which totals to 9. Can you please help
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If first is 8 and last can be either of 3,5,7,9 (odd numbers only) so in middle it will have only 8 numbers left to be filled out of 0-9 From your choices above (then middle is 0,1,2,3,4,5,6,7,9 - which totals to 9), you have taken all the 4 odd numbers (3,5,7,9) at same time. Which cannot be possible.
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Constraints :- 1. Three digit number 2. Starting with 8 or 9 3. No repetition 4. Odd When 8 is first digit:- 1. Second number even (0,2,4,6), third number (1,3,5,7,9)- total cases (4*5 =20) 2. Second number odd (1,3,5,7,9), third number (odd numbers -1=4) - total cases (5*4=20) When 9 is first digit:- 3. Second number even (0,2,4,6,8), third number odd (1,3,5,7)- total cases (5*4 = 20) 4. Second number is odd (1,3,5,7), third number odd (odd numbers -1 =3) - total cases (4*3 = 12) Total number of possible arrangements = 20+20+20+12 = 72 This is lengthy procedure and is risky because you may miss one case & answer will be completely different , IMO direct application of P&C theory will be better approach under time constraints .
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8 _ _
You need to pick 2 numbers from set { 0,1,2,3,4,5,6,7,9}. Hence 9C2 = 36. But u can order the 2 numbers around, so 36*2 = 72
9 _ _
You need to pick 2 numbers from set { 0,1,2,3,4,5,6,7,8}. Hence 9C2 = 36. But u can order the 2 numbers around, so 36*2 = 72
But you only need odd numbers so (72 + 72)/2 = 72.
Answer is C.
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I hope my approach is correct anyway; Let N = total no of digits with different individual digits N= 2x9x8 = 144 no of even = no of odds = 72. (D)
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I just used slot method: 2 choices for the hundreds digit (8 or 9), 9 choices for the tens digit (0,1,2,3,4,5,6,7, 8 or 9), and 4 choices for the tens digit (1 or 9, 3 or 9, 5 or 9 , 7 or 9) = 2 x 9 x 4 = 72.
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Between 800 and 899 -- (1)*(8)*(5) = 40 (units place -- 1,3,5,7,9)
Between 900 and 999 -- (1)*(8)*(4) = 32 (units place -- 1,3,5,7)
72
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first digit can be selected from (8,9) in 2 ways second distinct digit can be selected from (0,1,2,3,4,5,6,7,8,9) in 9 ways since one digit is already selected Third distinct digit can be selected from (0,1,2,3,4,5,6,7,8,9) in 8 ways since two digits are already selected . So total number of ways digits can be slected is 2*8*9 But we need to consider odd number of digits so (2*8*9)/2=72 SO...ANSWER is C
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k8llykh wrote: How many odd three-digit integers greater than 800 are there such that all their digits are different? (A) 40 (B) 56 (C) 72 (D) 81 (E) 104 Source: GMAT Club Tests - hardest GMAT questions Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker. For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part: For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3-digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)-5= 45-5= 40 Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4) - 4 = 36 - 4 = 32 possibilites. Add the 2 together, and you get 40+32=72. The answer! Hope that helps!! 1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. Total for this range: 1*5*8 = 40. In the range 900 - 999:1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. Total for this range: 1*4*8 = 32. Total: 40+32 = 72. Answer: C.
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Slot method. If hundreds digit is 8, there are 5 options to make last digit (ones digit) odd. That leaves 8 options for tens digits. Total is 1*8*5=40 If hundreds digit is 9, there are 4 options to make last digit (ones digit) odd. That leaves 8 options for tens digits. Total is 1*8*4=32 Total= 72 Answer is C. Cheers, Der alte Fritz.
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I just culculated (2*9*8)/2 = 72
(2*9*8) to get all nummers with all their digits different from 800 to 999
and than divided it by 2 to get only the odd numbers
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k8llykh wrote: How many odd three-digit integers greater than 800 are there such that all their digits are different? (A) 40 (B) 56 (C) 72 (D) 81 (E) 104 Source: GMAT Club Tests - hardest GMAT questions Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker. For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part: For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3-digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)-5= 45-5= 40 Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4) - 4 = 36 - 4 = 32 possibilites. Add the 2 together, and you get 40+32=72. The answer! Hope that helps!! good problem...every time i get such problems as these, i try to solve in conservative(old manual) method.....i should try to find the right method which is also the faster method....ANSWER is 72 and no doubt abt it.
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Bunuel wrote: k8llykh wrote: How many odd three-digit integers greater than 800 are there such that all their digits are different?
(A) 40
(B) 56
(C) 72
(D) 81
(E) 104
Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker.
For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part:
For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3-digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)-5= 45-5= 40
Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4) - 4 = 36 - 4 = 32 possibilites.
Add the 2 together, and you get 40+32=72. The answer!
Hope that helps!! In the range 800 - 900:1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. Total for this range: 1*5*8 = 40. In the range 900 - 999:1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. Total for this range: 1*4*8 = 32. Total: 40+32 = 72. Answer: C. Hi Geek, I wonder how u r solving all questions with proper explanations... HAts off for tat... How u r solving by these strategy? Does any math strategy is there for this types of problem and could you pls help me to read those things.... In the range 800 - 900:1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. Total for this range: 1*5*8 = 40. In the range 900 - 999:1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. Total for this range: 1*4*8 = 32. Cheers, Sharbu
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I am kicking myself for getting this problem wrong on account of a really bad silly mistake. I missed reading the "odd" in the question stem and ended up with getting an answer not in the given choices. The good thing is that I knew how to approach it using the combinatrics method but on account of poor reading of the question stem I ended up choosing a wrong answer. Bunuel's way of selecting the 3rd digit after selecting the first (8/9) was particularly helpful and quick. Missed this question only because of not reading with full concentration. Real bad mistake. 
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