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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!

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.

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

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

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. _________________

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 . _________________

Sun Tzu-Victorious warriors win first and then go to war, while defeated warriors go to war first and then seek to win.

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) _________________

KUDOS me if you feel my contribution has helped you.

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.

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 _________________

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.

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

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. _________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

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.

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. _________________

My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com

Hi everyone, most fastest way to solve this question is

8_ _, different odd no. will be 8 _ _. first digit is 8, for odd number third digit will be (1,3,5,7,9) i.e. 5 possibility 2nd digit we can fill from (0-9), since we are looking for different digit so choice for 2nd digit will 10(0-9) - 1st digit-2nd digit with 8 possibility 8 _(8) _(5) = 8*5 = 40

similarly for 9 _ _

for 3rd digit choices are {1,3,5,7} for 2nd digit choices are 10 - 1st digit-3rd digit = 8

so 8*4=32

so total = 72 different odd digits

Another simple wat to solve is b/w 800-900 we have 50 odd no. now we will substract the odd no. like {11,33,55,77,99}= 5 and 881,83,85,87,89 =5 i.e. 50-5-5 =40

b/w 900-999 we have 50 odd no substract {11,33,55,77,99}=5 and 991,93,95,97,99 = 5 and like 919,929 .....989=8 50-5-5-8 = 32 so different odd number will be 40+32 = 72