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is something wrong with this question? I think the total number of such digits is 80. It cannot be 72. Aren't all the solutions posted above counting numbers with similar digits? The questions states how many odd numbers with different digits.

There will be 45 numbers starting with 8 and 35 starting with 9. This gives a total of 80?

How many odd three-digit integers greater than 800 are there such that all their digits are different?

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.

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'm still not getting the logic..i find myself convincing myself of different things for examplle

for 800s first digit can be 8 only 2nd digit can be 0,1,2,3,4,5,6,7,9 3rd digit can be 1,3,5,7,9 1x9x5

for 900s 1st 9 2nd 0,1,2,3,4,5,6,7,8 3rd 1,3,5,7 1x9x4

how is the answer not 45 + 36 ?

This approach is not correct. If you say that (in the range 800-900) for the second digit you have 9 choices, then for the third digit you'll have sometimes 5 choices (in case you choose even for the second) and sometimes 4 choices (in case you choose odd for the second), so you can not write 1*9*5.

Correct way would be to count the choices for the third digit first - 5 choices (1, 3, 5, 7, 9) and only then to count the choices for the second digit - 8 choices (10-first digit-third digit=8) --> 1*5*8=40.

Similarly for 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.
_________________

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

Answer is D, between 801 to 999 total 199 numbers are there. a. Total odd numbers are(199/2+1) 100, because first and last numbers are odd, b. we have number from like (881, 883, 885, 887 & 889) - 5 numbers repeating 8 two times c. Like 9 repeating another 5 Numbers(991, 993, 995, 997 & 999) d. We have number like 811, 833, 855, 877 & 899, 5 Numbers e. We have another numbers like 911, 933, 955, 977 - numbers(999 already removed) Total numbers is = 100-5-5-5-4 = 81 numbers

Please correct me if I am wrong. Please remove the question if i am correct. Save precious time of aspirants by removing useless questions

When 100's digit is 8 ...the no. of digits is 8X5 =40 No . of cases when units digit is 1= 8 Now, there are 5 odd digits that can appear in units digits . therefore 8 X 5 =40

100's digit is 9 ...the no. of digits is 8 X 4 = 32 In this case the odd no. 9 is in 100'2 digit. Therefore it is not available for units digit. Hence the no. of add distinct digits is 8X4 =32

100s digit can be 8 or 9....hence total 2. 10's digit can be any number from 0 to 9 but excluding the digit that has been selected for 100s digit....hence a total of 9 ways.

unit digit can be 1,3,5,7 or 9....but 9 would be already selected previously, hence a total of 4.

Its 72. Odd nubmer in the last digit- 1, 3, 5, 7, 9 for the 800 series. Since 3# are different, 0, 2, 4, 6, 1,3,5,7,9 can be the middle digit. For the 800series, there will be 5x4+5x4=40 numbers

Odd number in the last digit- 1, 3, 5, 7 for the 900 series. Since 3# are different, 2, 4, 6, 8,1,3,5,7 can be the middle digit. For the 900series, there will be 4x5=4x3=32 numbers.

How many odd three-digit integers greater than 800 are there such that all their digits are different?

* 40 * 56 * 72 * 81 * 104

NUMBERS are 8XY and 9XY:

a. for 8XY:

X can odd or even integer other than 8 for integers between 800 and 900:

if X is even, y can be all 5 odd integers. so no of ways = 4*5 = 20 if X is odd, y can be all remaining 4 odd integers. so no of ways = 5*4 = 20

b. for 9XY:

X can odd or even integer other than 9 for integers between 900 and 1000:

if X is even, y can be all 4 odd integers except 9. so no of ways = 5*4 = 20 if X is odd, y can be all remaining 3 odd integers. so no of ways = 4*3 = 12

is something wrong with this question? I think the total number of such digits is 80. It cannot be 72. Aren't all the solutions posted above counting numbers with similar digits? The questions states how many odd numbers with different digits.

There will be 45 numbers starting with 8 and 35 starting with 9. This gives a total of 80?

Let the 1st digit be 8. The second digit can be either odd (5 options) or even (4 options). The third digit has 4 options (if the 2nd is odd) and 4 options (if the 2nd is even). So you get: 1x5x4 + 1x4x5 = 40

Let the 1st digit be 9. The second can be either odd (4 options) or even (5 options). The third digit has 3 option (if the 2nd is odd) and 4 options (if the 2nd is even). So you get: 1x4x3 + 1x5x4 = 32

Re: How many odd three-digit integers greater than 800 are there [#permalink]

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Re: How many odd three-digit integers greater than 800 are there [#permalink]

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13 Nov 2016, 02:29

study 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

Answer is D, between 801 to 999 total 199 numbers are there. a. Total odd numbers are(199/2+1) 100, because first and last numbers are odd, b. we have number from like (881, 883, 885, 887 & 889) - 5 numbers repeating 8 two times c. Like 9 repeating another 5 Numbers(991, 993, 995, 997 & 999) d. We have number like 811, 833, 855, 877 & 899, 5 Numbers e. We have another numbers like 911, 933, 955, 977 - numbers(999 already removed) Total numbers is = 100-5-5-5-4 = 81 numbers

Please correct me if I am wrong. Please remove the question if i am correct. Save precious time of aspirants by removing useless questions