GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Nov 2019, 01:12 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Of the three-digit positive integers that have no digits

Author Message
TAGS:

### Hide Tags

Manager  Joined: 10 Oct 2005
Posts: 103
Location: Hollywood
Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

2
25 00:00

Difficulty:   55% (hard)

Question Stats: 66% (02:09) correct 34% (01:56) wrong based on 783 sessions

### HideShow timer Statistics

Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?

A. 24
B. 36
C. 72
D. 144
E. 216

_________________
The GMAT, too tough to be denied.
Beat the tough questions...
Math Expert V
Joined: 02 Sep 2009
Posts: 59239
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

2
10
Baten80 wrote:
many different ways to solve the problem. Is there any unique/good/universal approach that can be applied to other similar problems.

I don' think that there is an unique or universal approach to such kind of problems. You should choose the one you are more comfortable with.

As for the above problem. One can also use combination approach.

Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216

We have a three digit integer: XXY.
Choosing the digit for X - 9 ways;
Choosing the digit for Y - 8 ways;
# of permutations of 3 digits in XXY - 3!/2! (as 2 X's are identical);

Total: 9*8*3!/2!=216.

Hope it helps.
_________________
Manager  Joined: 04 Jan 2006
Posts: 205
Re: PS - Digits (Very Tough one)  [#permalink]

### Show Tags

1
4
TOUGH GUY wrote:
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?

Assume: a,b,c is the digit and a,b,c not= 0

Thus a,b,c could be 1, 2, 3, 4, 5, 6, 7, 8, and 9

Count the possible way to get three digit number abc.

a = b, and c must be different from a, b
Thus, there are 3 possible ways of digit arrangement: aac, aca, caa

Case I: aac
=> (digit 1st) x (digit 2nd) x (digit 3rd)
=> 9 x 1 x 8 {pick any number from group = 9 possible ways} x {pick number the same as the first pick = 1 way} x {pick any number from the rest = 8 possible ways}

= 9 x 1 x 8 = 72 possible ways

Case II: aca
=> same as case I you have 72 possible ways

Case III: caa
=> same as case I you have 72 possible ways

total of this set of number = 72 + 72 + 72 = 216

(I got this question the first time 72 but I found that is question needs a little more work to do )
##### General Discussion
Director  Joined: 10 Oct 2005
Posts: 555
Re: PS - Digits (Very Tough one)  [#permalink]

### Show Tags

1
TOUGH GUY wrote:
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?

A. 24
B. 36
C. 72
D. 144
E. 216

three-digit positive integer where two digits that are equal to each other and the remaining digit different from the other two
A B C -three digits
possible combinations
AAC
ACA
CAA
1-1-(2,3,4,5,6,7,8,9) total 8 combinations
1-((2,3,4,5,6,7,8,9)-1 total 8 combinations
(2,3,4,5,6,7,8,9)-1,1 total 8 combinations
3*8=24
and so on

we have 9 digits with 24 possible comb for each
total 9*24=216
_________________
IE IMBA 2010
Director  Joined: 24 Aug 2006
Posts: 558
Location: Dallas, Texas

### Show Tags

Zero can't be a digit so the not equal can be selected 9 ways and equal can be selected 8 ways or vice versa.

If Number is XXY then there are 8*9=72 possibilities
If Number is XYX then there are 8*9 =72 possibilities
If Number is YXX then there are 8*9 =72 possibilities
Total 216
_________________
"Education is what remains when one has forgotten everything he learned in school."
Senior Manager  S
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 416
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

many different ways to solve the problem. Is there any unique/good/universal approach that can be applied to other similar problems.
_________________
Manager  Status: exam is close ... dont know if i ll hit that number
Joined: 06 Jun 2011
Posts: 126
Location: India
GMAT Date: 10-09-2012
GPA: 3.2
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

good question...

good explanation too..
_________________
just one more month for exam...
SVP  Joined: 06 Sep 2013
Posts: 1553
Concentration: Finance
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

Bunuel wrote:
Baten80 wrote:
many different ways to solve the problem. Is there any unique/good/universal approach that can be applied to other similar problems.

I don' think that there is an unique or universal approach to such kind of problems. You should choose the one you are more comfortable with.

As for the above problem. One can also use combination approach.

Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216

We have a three digit integer: XXY.
Choosing the digit for X - 9 ways;
Choosing the digit for Y - 8 ways;
# of permutations of 3 digits in XXY - 3!/2! (as 2 X's are identical);

Total: 9*8*3!/2!=216.

Hope it helps.

Is it possible to solve the inverse? Say TOTAL - (Cases were ALL are equal ) - (Cases were ALL are different)?

I'm trying to take a stab at it but I'm not able to come with the answer

Is it 9^3 - (9*3) - (9*8*7) ?

Thanks
Cheers
J
Math Expert V
Joined: 02 Sep 2009
Posts: 59239
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

1
jlgdr wrote:
Bunuel wrote:
Baten80 wrote:
many different ways to solve the problem. Is there any unique/good/universal approach that can be applied to other similar problems.

I don' think that there is an unique or universal approach to such kind of problems. You should choose the one you are more comfortable with.

As for the above problem. One can also use combination approach.

Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216

We have a three digit integer: XXY.
Choosing the digit for X - 9 ways;
Choosing the digit for Y - 8 ways;
# of permutations of 3 digits in XXY - 3!/2! (as 2 X's are identical);

Total: 9*8*3!/2!=216.

Hope it helps.

Is it possible to solve the inverse? Say TOTAL - (Cases were ALL are equal ) - (Cases were ALL are different)?

I'm trying to take a stab at it but I'm not able to come with the answer

Is it 9^3 - (9*3) - (9*8*7) ?

Thanks
Cheers
J

Almost nailed it! The # of three-digit integers that have all digits equal is just 9, not 9*3:
111,
222,
...
999.

So, {Total} - {all equal} - {all distinct} = (9*9*9) - 9 - (9*8*7) = 216.

Hope it's clear.
_________________
Current Student Joined: 08 Jun 2013
Posts: 31
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

1
Digits can be in the form ABB, BBA, BAB
For ABB - 9 options for A(1-9) and 8 options for B(2-9), since taking B as 1 will give 111 which is not required by question
hence no. of combinations possible 9*8 = 72
similarly for BBA- 9*8=72
and BAB - 9*8 =72
total = 72+72+72 = 216
Manager  B
Joined: 16 Mar 2016
Posts: 121
Location: France
GMAT 1: 660 Q47 V33 GPA: 3.25
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

We want a number like : XXY, X and Y being digits from 1 to 9.
To determine the number of possibilities, we can use the fundamental counting principle.

1/ Number of possibilities of X : 9
2/ Number of possibilities of Y : 8
3/ Number of ways to arranges XXY : 3!/2! = 3

So, 9*8*3 = 216
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2977
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

TOUGH GUY wrote:
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?

A. 24
B. 36
C. 72
D. 144
E. 216

Let, here are three places that we have to fill by digits and two of them must be filled by same digit

_ _ _

There are three cases

XXY ===> Total ways = 9*8 = 72 Numbers (X can be chosen 9 ways 1-9, and Y in another 8 ways)
XYX ===> Total ways = 9*8 = 72 Numbers (X can be chosen 9 ways 1-9, and Y in another 8 ways)
YXX ===> Total ways = 9*8 = 72 Numbers (Y can be chosen 9 ways 1-9, and Y in another 8 ways)

Total Numbers = 3*72 = 216

_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Intern  B
Joined: 31 Oct 2016
Posts: 5
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

Hi!

Can someone explain why the answer can't work using combination? For this example, I'm looking for the number of variations of AAB. I did 9 pick 2 (since you're using two digits and there's a total of 9 digits you can pick from). Then I multiplied 9 pick 2 (36) by 3!/2! which equals 108.
Math Expert V
Joined: 02 Sep 2009
Posts: 59239
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

1
GMATKAY22 wrote:
Hi!

Can someone explain why the answer can't work using combination? For this example, I'm looking for the number of variations of AAB. I did 9 pick 2 (since you're using two digits and there's a total of 9 digits you can pick from). Then I multiplied 9 pick 2 (36) by 3!/2! which equals 108.

9C2 = 36 is the number of pairs of different digits from 9 digits: (1, 2); (1, 3); (1, 4); ...; (8, 9).

Now, one pair, say (1, 2), can give you TWO types of numbers: with two 1 and one 2 and two 2's and one 1: 112 (which in turn gives 112, 121, 211) and 221 (which in turn gives 221, 212, 122). So, you should multiply 9C2*3!/2! further by 2: 9C2*3!/2!*2 = 216.

Hope it's clear.
_________________
GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4076
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

Top Contributor
TOUGH GUY wrote:
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?

A. 24
B. 36
C. 72
D. 144
E. 216

Take the task of creating suitable 3-digit numbers and break it into stages.

Stage 1: Select the single digit that will be different from the other 2 digits
We can choose any of the following 9 digits: 1, 2, 3, 4, 5, 6, 7, 8 or 9
So, we can complete stage 1 in 9 ways

Stage 2: Choose where that single digit (selected above) will be placed
There are 3 places (hundreds position, tens position or units position) where we can place this digit.
So, we can complete stage 2 in 3 ways.

At this point, we have 2 spaces left, and these spaces will be filled by the same digit.

Stage 3: Select the digit to occupy those two remaining spaces
There are 8 remaining digits from which to choose, so we can complete this stage in 8 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create the required 3-digit number) in (9)(3)(8) ways (= 216 ways)

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

RELATED VIDEOS

_________________
Intern  B
Joined: 19 Aug 2018
Posts: 4
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

Here 2 places can be selected in 3c2 ways and we have 9 options for first digit and 1 for the other similar digit
3rd different digit can be selected in 8 ways
so 3c2*9*1*8=216
Senior Manager  G
Joined: 04 Aug 2010
Posts: 493
Schools: Dartmouth College
Re: Of the three-digit positive integers that have no digits  [#permalink]

### Show Tags

TOUGH GUY wrote:
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?

A. 24
B. 36
C. 72
D. 144
E. 216

Integers with exactly 2 digits the same = Total integers - Integers with all 3 digits the same - Integers with all 3 digits different.

Total integers:
Number of options for the hundreds digit = 9. (Any digit but 0.)
Number of options for the tens digit = 9. (Any digit but 0.)
Number of options for the units digit = 9. (Any digit but 0.)
To combined these options, we multiply:
9*9*9.

Integers with all 3 digits the same:
111, 222, 333, 444, 555, 666, 777, 888, 999.
Number of options = 9.

Integers with all 3 digits different:
Number of options for the hundreds digit = 9. (Any digit but 0.)
Number of options for the tens digit = 8. (Any digit 1-9 other than the digit already used.)
Number of options for the units digit = 7. (Any digit 1-9 other than the two digits already used.)
To combine these options, we multiply:
9*8*7.

Thus:
Integers with exactly 2 digits the same = (9*9*9) - 9 - (9*8*7) = 9(81-1-56) = 9(24) = 216.

_________________
GMAT and GRE Tutor
Over 1800 followers
GMATGuruNY@gmail.com
New York, NY
If you find one of my posts helpful, please take a moment to click on the "Kudos" icon.
Available for tutoring in NYC and long-distance. Re: Of the three-digit positive integers that have no digits   [#permalink] 11 Jan 2019, 13:32
Display posts from previous: Sort by

# Of the three-digit positive integers that have no digits  