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

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Math Expert
Joined: 02 Aug 2009
Posts: 8609
Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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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

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

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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.
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Posts: 8609
Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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04 Nov 2017, 07:36
1
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
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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

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

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31 Jan 2018, 19:51
Hi All,

You'll likely find it easiest to handle this calculation in 'pieces'

This question lays out the following restrictions:
1) 3-digit numbers greater than 700
2) All digits are NON-0 and DIFFERENT
3) The number must be ODD.

1st digit is 7 = 1 option
3rd digit must be ODD, but NOT 7 = 4 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(4)(7) = 28 numbers in the 700s

Now, the 800s...

1st digit is 8 = 1 option
3rd digit must be ODD = 5 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(5)(7) = 35 numbers in the 800s

Finally, the 900s; the math here works the same as the 700s...

1st digit is 9 = 1 option
3rd digit must be ODD, but NOT 9 = 4 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(4)(7) = 28 numbers in the 900s

28+35+28 = 91

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

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18 Oct 2018, 20:54
Type-I: 7AB
Type-II: 8CD
Type-III- 9EF

Type-I = 1* A * B( Odd=1,3,5,7,9. Minus 7= 1,3,5,9= 4)
So, 1 * A * 7 (A is not 7 or B= 1,2,3,4,5,6,7,8,9= 7 and B= 7)
Now Type-I is 1*7*4

Type-II = 1 * C * D (D= odd= 1,3,5,7,9 = 5/ C= 1,2,3,4,5,6,7,8,9 minus 8 and C= 7)
So, 1* 7* 5

Type-III is the same as Type-I

Type-I: 1*7*4= 28
Type-II: 1*7*5=35
Type-III- 1*7*4=28
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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03 Sep 2019, 18:55
I looked at each set of "hundreds" in the range of numbers and went from there:
Range: 700-999

7 _ _ <--4 numbers: 1,3,5,9 can satisfy the odd constraint
7 _ <--(9-2 = 7 numbers can satisfy since the last digit and first digit need to be accounted for) _
For the 700 set, 7*4 =28 will give the possible integers within that range

Do the same for 8 and 9.
8 _ _ <--(1,3,5,7,9) Notice since 8 is not an odd number, the 7 will get added into the set of numbers that can take the last unit's place and the 4 becomes 5. For the middle digit, the 7 remains, since we still have 2 digits that cannot be repeated.
For the 800 set, 7*5=35

900s, will be the same as 700s. 7*4=28

Adding the 3 values resulting from each set's combination will give the total: 91 (28 + 28 + 35).
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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03 Sep 2019, 23:01
This is a very good question on Permutations involving rearrangement of digits to form numbers.

In most questions on permutations with digits to form numbers, it’s better to stick to the fundamental principle of counting rather than resorting to the usage of $$n_P_r$$ formula. This is because, when you stick to the multiplication principle/addition principle, you’ll be able to deal with different cases more efficiently.

On the other hand, if you are dealing with permuations of people or alphabets, the $$n_P_r$$ formula works better.

The questions says that the positive integers should be made of non-zero digits which are distinct. Also, the numbers should be greater than 700 and should be odd.This tells us the following things:

We can only use the digits from 1 to 9
We cannot have numbers like 111, 122 etc in our list.
We need to form numbers which look like 789, 827, 913 etc.,

The above is how this question needs to be visualized, because it will tell you what is allowed and what is not. This will then help you to tailor your approach accordingly.
Let us now break this down into three cases.

Case 1: Three digit odd numbers starting with 7.

Very often, students fall into the trap of assuming that the tens digit can be filled in 8 ways because there are 8 digits available. This is mainly due to the fact that they would have got accustomed to the process of filling up the digits from left to right, in most permutation questions related to numbers. This is not correct.

Always bear in mind that you have to deal with the constraints first. Here, the constraint is that the number that we form should be an odd number. A number is an odd number when it has 1, 3, 5, 7 or 9 in the units place.

Since the number 7 has already been used to fill the hundred’s place, we cannot fill the units place with 7. This means that there are four ways of filling the units place.
Once this is done, the tens place can now be filled in 7 ways, since 2 of the 9 available digits have already been used up.

For each way of filling the unit’s place, we have 7 ways of filling the ten’s place. Therefore, for 4 ways of filling the unit’s place, there will be 28 ways of filling the unit’s and the ten’s digits and so, 28 3-digit numbers are possible which are odd and greater than 700.

Case 2 : Three digit odd numbers starting with 9.

This case is exactly the same as the one with numbers starting with 7. The only exception here is that, we cannot use the digit 9 in the unit’s place, but, we can use the other 4 viz 1, 3, 5 and 7.
Therefore, we can obtain 28 3-digit numbers which are odd and greater than 900 (and hence greater than 700).

Case 3: Three digit numbers starting with 8.

IN this case, we can fill up the unit’s place with any of the 5 digits i.e. 1,3,5, 7 or 9. Therefore, in this case, we will have a total of 35 numbers.

Hence, the total number of odd numbers greater than 700 = 28 + 28 + 35 = 91.
The correct answer option is B.

As you may have observed, there’s not even a single place where we used any formula to solve the question. It was based purely on logic and basic counting principles. This is why P&C questions can be a double edged sword sometimes, but there’s no doubt that they are always fun to solve.

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

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20 Sep 2019, 07:45
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

Asked: Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

Let 3-digit positive integer be xyz
z = {1,3,5,7,9}
y = {1,2,3,4,5,6,7,8,9}
x = {7,8,9}

Since all 3 digits are different

Number of such 3-digit integers whose three digits are all different and nonzero, how many are odd integers greater than 700
OOO
2*4*3 = 24
OEO
2*4*4 = 32
EOO
1*5*4= 20
EEO
1*3*5 = 15

Total numbers = 24 + 32 + 20 + 15 = 91

IMO B
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Re: Of the three-digit positive integers whose three digits are all differ   [#permalink] 20 Sep 2019, 07:45

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