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Of the three-digit positive integers whose three digits are all differ

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Joined: 02 Aug 2009
Posts: 5344

Kudos [?]: 6111 [0], given: 121

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

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New post 04 Nov 2017, 04:08
pra1785 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

[Reveal] Spoiler:
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.


Hi,

I am a little confused with the explanation. For number of combinations for type 2: why is it 1*7*5
Don't we have to make sure that all numbers are different, non zero and odd in type 2 as well?

What am I missing?


Hi..
The last digit is odd and can be any of the five digits- 1,3,4,7,9
In type 2, the Hundreds digit is 8, so units digit can be any of 5 odd digits.
But in other type, the Hundreds digit is an odd integer-7 or 9, so units digit can be any of the remaining 4 as all digits are different
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6111 [0], given: 121

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Joined: 15 Aug 2016
Posts: 42

Kudos [?]: 1 [0], given: 74

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

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New post 04 Nov 2017, 07:31
I am confused.
7AB
8BC
9CD.

Considering 7AB.
7 is fixed and is odd. Thus B will have 4 possibilities. But I think A has 8 possibilities after we fill one value with A.
Let say we put 1 for B.
A can have 0,2,3,4,5,6,8,9 i.e 8 possibilities.
So for 1st case its 1*8*4=32
2nd case is 1*8*5=40
3rd case is same as 1st i.e 1*8*4=32
=>32+40+32=104.
Please tell me where I am going wrong.

Kudos [?]: 1 [0], given: 74

Expert Post
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Math Expert
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Joined: 02 Aug 2009
Posts: 5344

Kudos [?]: 6111 [1], given: 121

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

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New post 04 Nov 2017, 07:36
1
This post received
KUDOS
Expert's post
Utkarsh KOhli wrote:
I am confused.
7AB
8BC
9CD.

Considering 7AB.
7 is fixed and is odd. Thus B will have 4 possibilities. But I think A has 8 possibilities after we fill one value with A.
Let say we put 1 for B.
A can have 0,2,3,4,5,6,8,9 i.e 8 possibilities.
So for 1st case its 1*8*4=32
2nd case is 1*8*5=40
3rd case is same as 1st i.e 1*8*4=32
=>32+40+32=104.
Please tell me where I am going wrong.


hi..

be careful while reading the statements. many times we all tend to overlook finer points..
same case here..
it says three digits are different and NON ZERO..
so coloured portion above will be 7 possiblities
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6111 [1], given: 121

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Joined: 24 Jun 2017
Posts: 119

Kudos [?]: 14 [0], given: 129

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

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New post 04 Nov 2017, 09:08
So we have 3 digits numbers starting with 7 , 8 and 9. No repetitive digits, no zeros, numbers should be odd so ending with odd digit
7AB, 8AB, 9AB where A even number (excluding 0) and B is odd number (excluding repetition)
even numbers are 2, 4, 6, 8
odd numbers are 1, 3, 5, ,7 ,9

So for 7AB 1*4*4 (as we have to exclude 7 from B) = 16
So for 8AB 1*3*5 (as we have to exclude 8 from A) = 15
So for 9AB 1*4*4 (as we have to exclude 9 from B) = 16

Now we can have 7BB type combinations well, where BB are both odd numbers
So for 7BB 1*4*3 (as we have to exclude 7 from B) = 12
So for 8BB 1*5*4 = 20
So for 9BB 1*4*3 (as we have to exclude 9 from B) = 12

Total 16+15+16+12+20+12=91

Kudos [?]: 14 [0], given: 129

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Joined: 15 Aug 2016
Posts: 42

Kudos [?]: 1 [0], given: 74

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

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New post 04 Nov 2017, 09:26
chetan2u wrote:
Utkarsh KOhli wrote:
I am confused.
7AB
8BC
9CD.

Considering 7AB.
7 is fixed and is odd. Thus B will have 4 possibilities. But I think A has 8 possibilities after we fill one value with A.
Let say we put 1 for B.
A can have 0,2,3,4,5,6,8,9 i.e 8 possibilities.
So for 1st case its 1*8*4=32
2nd case is 1*8*5=40
3rd case is same as 1st i.e 1*8*4=32
=>32+40+32=104.
Please tell me where I am going wrong.


hi..

be careful while reading the statements. many times we all tend to overlook finer points..
same case here..
it says three digits are different and NON ZERO..
so coloured portion above will be 7 possiblities


Hi Chetan.
Thank you for pointing it out. Kudos to you.

Kudos [?]: 1 [0], given: 74

Re: Of the three-digit positive integers whose three digits are all differ   [#permalink] 04 Nov 2017, 09:26

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