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

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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
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To find odd integers greater than 700:

In 700s:
Units can be any of 4 (1,3,5,9)
Hundreds is obviously just 1 way (7)
So, tens can be any of the remaining 7.
So, total = 4*1*7 = 28

In 800s:
Units can be any of 5 (1,3,5,7,9)
Hundreds is obviously just 1 way (8)
So, tens can be any of the remaining 7.
So, total = 5*1*7 = 35

In 900s:
Units can be any of 4 (1,3,5,7)
Hundreds is obviously just 1 way (9)
So, tens can be any of the remaining 7.
So, total = 4*1*7 = 28

So, grand total = 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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We know the integers must be greater than 700 and they must be odd; we can divide them into 6 groups: 7EO, 7OO, 8EO, 8OO, 9EO and 9OO where E denotes an even digit and O an odd digit. Let’s determine the number of integers in each group.

7EO:
The hundreds digit is 7, so it’s only 1 choice. Since all the digits are nonzero, E can be 4 numbers: 2, 4, 6 or 8. Since the digits are all different, O can be 4 numbers: 1, 3, 5 or 9. Therefore, there are 1 x 4 x 4 = 16 such integers.

7OO:
Since all the digits are different, O can only be 4 numbers: 1, 3, 5 or 9. However, the second O can only be 3 numbers since it cannot be the same as the first O. Therefore, there are 1 x 4 x 3 = 12 such integers.

8EO:
The hundreds digit is 8, so there is only 1 option. Since all the digits are different and nonzero, E can only be 3 numbers: 2, 4, or 6. Since the last digit is O, an odd digit, O can any of the 5 odd digits: 1, 3, 5, 7 or 9. Therefore, there are 1 x 3 x 5 = 15 such integers.

8OO:
The first O can be any of the 5 odd digits: 1, 3, 5, 7 or 9. However, the second O can only be 4 numbers since it cannot be the same as the first O. Therefore, there are 1 x 5 x 4 = 20 such integers.

9EO:
This is analogous to 7EO, so there should be 16 such integers.

9OO:
This is analogous to 9EO, so there should be 12 such integers.

Therefore, there are 16 + 12 + 15 + 20 + 16 + 12 = 91 odd integers that are greater than 700 where all digits are different and nonzero.

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

Hi,
since you want to know where you have gone wrong in your method..
hundreds digit - 3-- correct
units digit - 4 -- correct but ONLY in two cases where hundreds digit is 7 and 9-- MEANS you will have to add later for hundreds digit as 8, since 8 will have 5 ODDs as units digit..
I assume you mean tens digit as 8..NO, we are looking for Non-zero, so 10 - 2( ONE of hundreds and ONE of units) - 1( zero) = 7..

Now our answer is -- 3*4*7 = 84..
But we have to add for 5th Odd in units place for 8 as hundred's digit = 1*7*1 = 7..

Total = 84+7 = 91
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
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Hi chetan2u,

I did this question from 2 approach. By 1 I am not getting the correct ans.
Let me tell you both of my approaches

Approach 1

For thousand's digit we have 3 options- 7,8 and 9
For Unit's digit we have 8 options - Excluding 0 and the first digit
For Hundered's digit we have 7 options - Excluding 0 ,first and second digit
So= 3x8x7= 168 (Include both Odd and Even terms)
So, 168/2= 84 which is not the answer

Approach 2

for 700-799= 1*7*4= 28
for 800-899= 1*7*5 = 35
for 900-999= 1*7*4= 28
28+35+28 = 91 which is correct answer

__________________________________________________-

But we have to add for 5th Odd in units place for 8 as hundred's digit = 1*7*1 = 7..

I see that you to added 7 to 84.

But I am still not getting, whats arong in my approach 1. or what extra numbers do I need to add in 84.

Thanks
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
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PrakharGMAT wrote:
Hi chetan2u,

I did this question from 2 approach. By 1 I am not getting the correct ans.
Let me tell you both of my approaches

Approach 1

For thousand's digit we have 3 options- 7,8 and 9
For Unit's digit we have 8 options - Excluding 0 and the first digit
For Hundered's digit we have 7 options - Excluding 0 ,first and second digit
So= 3x8x7= 168 (Include both Odd and Even terms)
So, 168/2= 84 which is not the answer

Approach 2

for 700-799= 1*7*4= 28
for 800-899= 1*7*5 = 35
for 900-999= 1*7*4= 28
28+35+28 = 91 which is correct answer

__________________________________________________-

But we have to add for 5th Odd in units place for 8 as hundred's digit = 1*7*1 = 7..

I see that you to added 7 to 84.

But I am still not getting, whats arong in my approach 1. or what extra numbers do I need to add in 84.

Thanks

Hi,

Approach 1

For thousand's digit we have 3 options- 7,8 and 9
For Unit's digit we have 8 options -Excluding 0 and the first digit
For Hundered's digit we have 7 options - Excluding 0 ,first and second digit
So= 3x8x7= 168 (Include both Odd and Even terms)
So, 168/2= 84 which is not the answer

you can HALF ONLY when the numbers of ODD and EVEN are same,..
here it is not the same..
UNITS digit can be any of 10, 0 to 9, but we are excluding 0.. so there are 5 ODD- 1,3,5,7,9- and 4 EVEN - 2,4,6,8..
Again when 7 and 9 are first digit , the numbers would be different and when it is 8, it will be different
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
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PrakharGMAT wrote:
Hi chetan2u,

Approach 1

For thousand's digit we have 3 options- 7,8 and 9
For Unit's digit we have 8 options - Excluding 0 and the first digit
For Hundered's digit we have 7 options - Excluding 0 ,first and second digit
So= 3x8x7= 168 (Include both Odd and Even terms)
So, 168/2= 84 which is not the answer

Thanks

Hi PrakharGMAT,
If you want to follow approach I..
first find for 7 and 9..
hundreds digit - 7 or 9- 2 ways
units digit - 1,2,3,4,5,6,8,7/9 - so 8 --- 4 are even and 4 are ODD..
tens digit - remaining 7..
total ways = 2*8*7 = 112..
half of these will be ODD so 112/2 = 56

Now if hundreds digit is 8..
units digit = 9 but ONLY 5 will be ODD
tens digit remaining 7..
so total = 1*7*5=35

Overall = 56+35 = 91..
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
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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

i tried to solve this way:

700-1000 there 300 numbers
300/2 odds so 150 numbers
710-720-730 .... 10 *3 =30 numbers so 150-30=120
than 701-702-703 ..... 10*3=30 numbers so 120-30=90
90 is close to 91.
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
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According to the constraints we have 3 options for the thousands digit (7, 8, 9), 9 options for the hundreds digit (except 0), and 5 options for the units digit (it needs to be an odd number) in total with additional constraint that all digits should be different.
Lets look at two possible cases.

First (thousands digit is odd)
2 x 7 x 4 = 56
___ ____ ____
(7,9) (9-2) (5-1)

Thousands digit is even
1 x 7 x 5 = 35
___ ____ ____
(8) (9-2) (5)

Total 56+35=91
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
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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

For the 700s:
_ _ _

I determined the possible options for each slot. The number of options for the first slot is 1 (7). For the second slot the number can be even or odd. Starting with the odd case, the number of options is 4 (1, 3, 5, 9). This means that the number of options available for the third slot is 3. 1x4x3=12.
Now, if the second slot is even, there are 4 options (0 cannot be counted). The number of options for the third slot is 4 (only one odd number is used up - by the first slot). 1x4x4=16.

12+16=28. This number of options is the same for the 900s. So, now we just need to find the number of options for the 800s:

The number of options for the first slot is 1 (8). For the second slot the number can be even or odd. Starting with the odd case, the number of options is 5 (1, 3, 5, 7, 9). This means that the number of options available for the third slot is 4. 1x5x4=20.
Now, if the second slot is even, there are 3 options (0 cannot be counted). The number of options for the third slot is 5 (no other odd numbers used up). 1x3x5=15.

20+15=35.

Total = 28+28+35=91.

Kudos if you agree! Please comment if you disagree with the method and/or have improvements.
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
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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

1. Units digit can have 4 values i.e., 1,3,5,9 and 1,3,5,7 when the hundreds digit is 7 and 9 resp and can have 5 values i.e, 1,3,5,7,9 when the hundreds digit is 8
2. Tens digit cannot have the value of unit and hundreds digit and cannot be 0. So can have 7 values
3. So when hundreds digit is 7 or 9, number of possibilities is 4*7 + 4*7 = 56 and when hundreds digit is 8, number of possibilities is 5*7=35
4. Total number of possibilities is 91
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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 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?
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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
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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]
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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Re: Of the three-digit positive integers whose three digits are all differ [#permalink]
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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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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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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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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.

Love your method, so clear and concise. One question though, why is the second slot 7? Since you don't have limitations other than the fact that you can't use 0 or 7 in the second slot, and can't be the same numbers as 1,3,5,9, shouldn't the number of choices in the second slot be 4? (from digit 0-9, there are 10 choices, minus 0, 7, 1,3,5,9, you have 4 choices left.)
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