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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
ENGRTOMBA2018 wrote:
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help



Please follow posting guidelines (link in my signatures). Do make sure to mention the correct title and put your analyses either in the next post or under spoilers.


As for the question, you need to find the number of combinations possible such that:

1. It is a 3 digit number > 700 (or between 701-999, inclusive).
2. All digits are different and NON ZERO.
3. The numbers must be ODD --> the last digit can be 1 of 1,3,5,7,9

Based on this, the numbers can be of the following 3 types:

Type 1: 7AB
Type 2: 8EF
Type 3: 9CD

For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF).

Number of combinations for type 1 : 1*7*4 = 28

Number of combinations for type 2: 1*7*5 = 35

Number of combinations for type 3 : 1*7*4 = 28

Thus, total numbers possible = 28+35+28 = 91.

B is thus the correct answer.

Hope this helps.


Why did you multiply with 7?
We have total 9 digits. One is used in hundreds place, then we must be left with 8 digits for tens place. Please help me understand where am I going wrong
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
adityaganjoo wrote:
ENGRTOMBA2018 wrote:
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help



Please follow posting guidelines (link in my signatures). Do make sure to mention the correct title and put your analyses either in the next post or under spoilers.


As for the question, you need to find the number of combinations possible such that:

1. It is a 3 digit number > 700 (or between 701-999, inclusive).
2. All digits are different and NON ZERO.
3. The numbers must be ODD --> the last digit can be 1 of 1,3,5,7,9

Based on this, the numbers can be of the following 3 types:

Type 1: 7AB
Type 2: 8EF
Type 3: 9CD

For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF).

Number of combinations for type 1 : 1*7*4 = 28

Number of combinations for type 2: 1*7*5 = 35

Number of combinations for type 3 : 1*7*4 = 28

Thus, total numbers possible = 28+35+28 = 91.

B is thus the correct answer.

Hope this helps.


Why did you multiply with 7?
We have total 9 digits. One is used in hundreds place, then we must be left with 8 digits for tens place. Please help me understand where am I going wrong


It is easier if you choose first the last digit and after that you choose the tens digit.
For example, in the 7XX case, if you want to do it choosing first the tens digits you have to differenciate two cases:
- ten digits is an even number: you have 4 options (2,4,6,8) and you leave 4 options for units digit (1,3,5,9): 1x4x4 = 16
- ten digits is an odd number: you have 4 options (1,3,5,9) and you leave only 3 options for units digit (you can not choose the same digit as in tens digit): 1x4x3 = 12

16+12 = 28

The result is the same but it is more complicated to calculate.
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
Hi,
We need 3 digit numbers greater than 700 which have all 3 different digits and number should be odd.
Let's fix hundreds places as 7,8 and 9, find EVEN/ODD combinations at tens and units place will be ODD-

7AB --> hundreds place is fixed, tens place can be even/odd and unit place will be odd
7EO- 1*4(2,4,6,8) * 4 (1,3,5,9) =16
7OO- 1* 4 (1,3,5,9) * 3 (remaining 3 digits from odd list after using 1 odd digit at Tens place) = 12

8CD-->hundreds place is fixed, tens place can be even/odd and unit place will be odd
8EO- 1* 3(2,4,6) * 5(1,3,5,7,9) = 15
8OO- 1* 5(1,3,5,7,9) * 4 (remaining 4 digits from odd list after using 1 odd digit at Tens place) = 20

9EF-->hundreds place is fixed, tens place can be even/odd and unit place will be odd
9EO- 1*4(2,4,6,8) * 4 (1,3,5,7) =16
9OO- 1* 4 (1,3,5,7) * 3 (remaining 3 digits from odd list after using 1 odd digit at Tens place) = 12

Add all values- 16+12+15+20+16+12 = 91
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
The key to solving the given equation is getting the combination right let us go
odd numbers 1,3,5,7,9-5
above 700= 7__=>7 and 4 therfore 28 numbers can be formed
above 800= 8__=>7and 5 therefore 35 numbers can be formed
above 900=9__=>7 and 4 therefore 28 numbers can be formed
total no of numbers = 91
Hence IMO B
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
ENGRTOMBA2018 wrote:
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help



Please follow posting guidelines (link in my signatures). Do make sure to mention the correct title and put your analyses either in the next post or under spoilers.


As for the question, you need to find the number of combinations possible such that:

1. It is a 3 digit number > 700 (or between 701-999, inclusive).
2. All digits are different and NON ZERO.
3. The numbers must be ODD --> the last digit can be 1 of 1,3,5,7,9

Based on this, the numbers can be of the following 3 types:

Type 1: 7AB
Type 2: 8EF
Type 3: 9CD

For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF).

Number of combinations for type 1 : 1*7*4 = 28

Number of combinations for type 2: 1*7*5 = 35

Number of combinations for type 3 : 1*7*4 = 28

Thus, total numbers possible = 28+35+28 = 91.

B is thus the correct answer.

Hope this helps.



Superbly explained , Thanks alot
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
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