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

It is currently 19 Sep 2018, 11:49

Close

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
Your Progress

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.

Close

Request Expert Reply

Confirm Cancel

How many 3 digit numbers can we make such that two of the digits are

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Manager
Manager
User avatar
Joined: 27 Nov 2006
Posts: 63
How many 3 digit numbers can we make such that two of the digits are  [#permalink]

Show Tags

New post 07 Jul 2018, 04:28
1
1
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

57% (01:30) correct 43% (01:40) wrong based on 121 sessions

HideShow timer Statistics

How many 3 digit numbers can we make such that two of the digits are same and non of the digits equal zero?

A. 60
B. 72
C. 150
D. 216
E. 280
Senior Manager
Senior Manager
User avatar
G
Joined: 14 Feb 2018
Posts: 379
Re: How many 3 digit numbers can we make such that two of the digits are  [#permalink]

Show Tags

New post 07 Jul 2018, 05:42
1
Following are the possibilities -

**# or *## or *#*
9*8 or 9*8 or 9*8 ways
i.e., 72*3 = 216 ways. Thus, D is the answer.

Posted from my mobile device
Manager
Manager
User avatar
B
Joined: 13 Feb 2018
Posts: 131
CAT Tests
Re: How many 3 digit numbers can we make such that two of the digits are  [#permalink]

Show Tags

New post 07 Jul 2018, 10:50
We have to consider several scenarios
1) xxy 9*1*8=72
2) xyx 9*8*1=72
3) yxx 8*9*1=72

3*72=216
Where x is the same digits and y is the different one

In My Opinion
Ans: D
Senior Manager
Senior Manager
User avatar
P
Joined: 14 Dec 2017
Posts: 471
Re: How many 3 digit numbers can we make such that two of the digits are  [#permalink]

Show Tags

New post 07 Jul 2018, 12:44
Since the digits are non zero, hence 9 digits {1,2,....9} available to chose from

First digit = 9C1 = 9
Second digit = same as first digit = 1
Third digit = 8C1 = 8

Arrangements possible = 3!/2! = 3


Hence total 3 digit such #'s possible = 9*1*8*3 = 216


Answer D.



Thanks,
GyM
_________________

New to GMAT Club - https://gmatclub.com/forum/new-to-gmat-club-need-help-271131.html#p2098335

Manager
Manager
User avatar
Joined: 27 Nov 2006
Posts: 63
Re: How many 3 digit numbers can we make such that two of the digits are  [#permalink]

Show Tags

New post 07 Jul 2018, 13:13
GyMrAT wrote:
Since the digits are non zero, hence 9 digits {1,2,....9} available to chose from

First digit = 9C1 = 9
Second digit = same as first digit = 1
Third digit = 8C1 = 8

Arrangements possible = 3!/2! = 3


Hence total 3 digit such #'s possible = 9*1*8*3 = 216


Answer D.
Thanks,
GyM

Thanks everyone for the detailed answer however I have the following few questions.

1. Why use 9C1 and 8C1? Usually when we use Fundamental Counting Principle, we simply put the # of possibilities at each step. If we didn't have the restriction that two digits are the same then we would have solved it by 9*8*7. Is combination generally hidden in FCP?

2. Also why did you say "Arrangements Possible = 3!/2!"? I thought the permutation took care of all arrangements when we compute 9*1*8 (XXY) - as the permutation definition says that if an event can occur in m ways and a second event in n ways then the two events can occur in m*n ways. When we have the general form XYZ, is 3!/3! hidden in the calculation?

Regards,
Senior Manager
Senior Manager
User avatar
P
Joined: 14 Dec 2017
Posts: 471
Re: How many 3 digit numbers can we make such that two of the digits are  [#permalink]

Show Tags

New post 09 Jul 2018, 23:11
computer-bot wrote:

Thanks everyone for the detailed answer however I have the following few questions.

1. Why use 9C1 and 8C1? Usually when we use Fundamental Counting Principle, we simply put the # of possibilities at each step. If we didn't have the restriction that two digits are the same then we would have solved it by 9*8*7. Is combination generally hidden in FCP?

Since we are choosing one digit at a time to fill a slot. Suppose you were given 3 different bowls & 9 different marbles and told that each bowl should have 2 marbles. You will do that in 9C2*7C2*5C2

2. Also why did you say "Arrangements Possible = 3!/2!"? I thought the permutation took care of all arrangements when we compute 9*1*8 (XXY) - as the permutation definition says that if an event can occur in m ways and a second event in n ways then the two events can occur in m*n ways. When we have the general form XYZ, is 3!/3! hidden in the calculation?

The keyword you are missing is "Independent" events.

Since all the slots (hundred's place, tens place & units place) are distinct, the repeated digit has three options to be in. When we have XXY arrangement, it means that we are selecting one place (either hundred's or tens place) for the repeated digit & arranging the other two places with 2 unique digits.

Consider Hundreds place for repeated digit is selected, we have 9 options for the tens Place, 8 options for the units place & 1 option for the hundred's place

For e.g. We select 1, we have arrangements,

When 1 is in hundred's place as 112, 113, 114, 115,.....119, 121, 131, 141,.....191 - 16 #'s

Now 1 in tens place, we have, 112, 113, 114,.....119, 911, 811, 711,....211 - 16#'s

Now 1 in units place, we have, 121, 131, 141,....191, 911, 811, 711,......211 - 16#'s

So we have Total Arrangements = 16 + 16 + 16 = 48, but as you can see half of which are duplicates. Hence we consider only 24 unique ones.

You can perform this for all the other remaining 8 digits, getting only 24 unique arrangements.




If you want to master Combinatorics, i suggest, if you have not already, then devote an obscene amount of time in assimilating the concepts explained by Karishma in the below article. Its a Combinatorics Bible.

https://gmatclub.com/forum/combinatoric ... l#p1579442


Hope that helps.


Thanks,
GyM
_________________

New to GMAT Club - https://gmatclub.com/forum/new-to-gmat-club-need-help-271131.html#p2098335

Intern
Intern
User avatar
B
Joined: 08 Jul 2016
Posts: 38
Location: Singapore
GMAT 1: 570 Q43 V25
GMAT 2: 640 Q42 V36
WE: Underwriter (Insurance)
Re: How many 3 digit numbers can we make such that two of the digits are  [#permalink]

Show Tags

New post 10 Jul 2018, 00:23
computer-bot wrote:
GyMrAT wrote:
Since the digits are non zero, hence 9 digits {1,2,....9} available to chose from

First digit = 9C1 = 9
Second digit = same as first digit = 1
Third digit = 8C1 = 8

Arrangements possible = 3!/2! = 3


Hence total 3 digit such #'s possible = 9*1*8*3 = 216


Answer D.
Thanks,
GyM

Thanks everyone for the detailed answer however I have the following few questions.

1. Why use 9C1 and 8C1? Usually when we use Fundamental Counting Principle, we simply put the # of possibilities at each step. If we didn't have the restriction that two digits are the same then we would have solved it by 9*8*7. Is combination generally hidden in FCP?

2. Also why did you say "Arrangements Possible = 3!/2!"? I thought the permutation took care of all arrangements when we compute 9*1*8 (XXY) - as the permutation definition says that if an event can occur in m ways and a second event in n ways then the two events can occur in m*n ways. When we have the general form XYZ, is 3!/3! hidden in the calculation?

Regards,


Hi mate,

U are getting too technical :)
For a 3 digit number, wherein u cannot have a zero and just 2 numbers repeated, make a slot - _ _ _
In first slot you can have any number from 1 to 9, so you have 9 options there. Second slot (the number being same as slot 1), you have only 1 option. Third slot can have any number other than the one in slot 1 and 2, giving you only 8 options. So you get 9*1*8 = 72

Note, you can have these numbers in 3 orders. Hence multiply 72 by 3.

Did that help? Is it any easier for you now?
Manager
Manager
User avatar
Joined: 27 Nov 2006
Posts: 63
Re: How many 3 digit numbers can we make such that two of the digits are  [#permalink]

Show Tags

New post 12 Jul 2018, 03:22
GyMrAT wrote:
If you want to master Combinatorics, i suggest, if you have not already, then devote an obscene amount of time in assimilating the concepts explained by Karishma in the below article. Its a Combinatorics Bible.

https://gmatclub.com/forum/combinatoric ... l#p1579442


Hope that helps.


Thanks,
GyM


Thanks. I haven't gone through it yet but I needed it.


Blueblu wrote:
Hi mate,

U are getting too technical :)
For a 3 digit number, wherein u cannot have a zero and just 2 numbers repeated, make a slot - _ _ _
In first slot you can have any number from 1 to 9, so you have 9 options there. Second slot (the number being same as slot 1), you have only 1 option. Third slot can have any number other than the one in slot 1 and 2, giving you only 8 options. So you get 9*1*8 = 72

Note, you can have these numbers in 3 orders. Hence multiply 72 by 3.

Did that help? Is it any easier for you now?



Blueblu... thanks for the response. What I don't understand is that if we had all three distinct digits XYZ, we would have simply multiplied by 9*8*7. But XYZ can also be arranged in the following 6 ways but we don't multiple X*Y*Z with 6.

XYZ
XZY
YXZ
YZX
ZXY
ZYX

Why is that so????
Intern
Intern
User avatar
B
Joined: 08 Jul 2016
Posts: 38
Location: Singapore
GMAT 1: 570 Q43 V25
GMAT 2: 640 Q42 V36
WE: Underwriter (Insurance)
Re: How many 3 digit numbers can we make such that two of the digits are  [#permalink]

Show Tags

New post 13 Jul 2018, 01:36
computer-bot wrote:
GyMrAT wrote:
If you want to master Combinatorics, i suggest, if you have not already, then devote an obscene amount of time in assimilating the concepts explained by Karishma in the below article. Its a Combinatorics Bible.

https://gmatclub.com/forum/combinatoric ... l#p1579442


Hope that helps.


Thanks,
GyM


Thanks. I haven't gone through it yet but I needed it.


Blueblu wrote:
Hi mate,

U are getting too technical :)
For a 3 digit number, wherein u cannot have a zero and just 2 numbers repeated, make a slot - _ _ _
In first slot you can have any number from 1 to 9, so you have 9 options there. Second slot (the number being same as slot 1), you have only 1 option. Third slot can have any number other than the one in slot 1 and 2, giving you only 8 options. So you get 9*1*8 = 72

Note, you can have these numbers in 3 orders. Hence multiply 72 by 3.

Did that help? Is it any easier for you now?



Blueblu... thanks for the response. What I don't understand is that if we had all three distinct digits XYZ, we would have simply multiplied by 9*8*7. But XYZ can also be arranged in the following 6 ways but we don't multiple X*Y*Z with 6.

XYZ
XZY
YXZ
YZX
ZXY
ZYX

Why is that so????


You are confusing distinct digits with the number of options. All of your digits X Y and Z are one of the digits from 1 to 9. Say you pick 678 and compare it with 876; in the first slot you put 9 which takes care of all 9 options including the digits 6,7 and 8 and so forth. This is why you don't multiply 9*8*7 with 3. I agree with GymArt - read Karishma's bible to get a clarity of your concepts.
Manager
Manager
User avatar
Joined: 27 Nov 2006
Posts: 63
Re: How many 3 digit numbers can we make such that two of the digits are  [#permalink]

Show Tags

New post 14 Jul 2018, 12:27
Blueblu wrote:
computer-bot wrote:
GyMrAT wrote:
If you want to master Combinatorics, i suggest, if you have not already, then devote an obscene amount of time in assimilating the concepts explained by Karishma in the below article. Its a Combinatorics Bible.

https://gmatclub.com/forum/combinatoric ... l#p1579442


Hope that helps.


Thanks,
GyM


Thanks. I haven't gone through it yet but I needed it.


Blueblu wrote:
Hi mate,

U are getting too technical :)
For a 3 digit number, wherein u cannot have a zero and just 2 numbers repeated, make a slot - _ _ _
In first slot you can have any number from 1 to 9, so you have 9 options there. Second slot (the number being same as slot 1), you have only 1 option. Third slot can have any number other than the one in slot 1 and 2, giving you only 8 options. So you get 9*1*8 = 72

Note, you can have these numbers in 3 orders. Hence multiply 72 by 3.

Did that help? Is it any easier for you now?



Blueblu... thanks for the response. What I don't understand is that if we had all three distinct digits XYZ, we would have simply multiplied by 9*8*7. But XYZ can also be arranged in the following 6 ways but we don't multiple X*Y*Z with 6.

XYZ
XZY
YXZ
YZX
ZXY
ZYX

Why is that so????


You are confusing distinct digits with the number of options. All of your digits X Y and Z are one of the digits from 1 to 9. Say you pick 678 and compare it with 876; in the first slot you put 9 which takes care of all 9 options including the digits 6,7 and 8 and so forth. This is why you don't multiply 9*8*7 with 3. I agree with GymArt - read Karishma's bible to get a clarity of your concepts.

On it :)

Thanks
Director
Director
User avatar
G
Joined: 20 Feb 2015
Posts: 728
Concentration: Strategy, General Management
Premium Member CAT Tests
Re: How many 3 digit numbers can we make such that two of the digits are  [#permalink]

Show Tags

New post 30 Jul 2018, 07:04
computer-bot wrote:
How many 3 digit numbers can we make such that two of the digits are same and non of the digits equal zero?

A. 60
B. 72
C. 150
D. 216
E. 280


step 1 .
two of the digits are same
assume we have 3 places _ _ _
no of ways we can select 2 out of these three to be the same is 3c2 = 3

step 2.
selecting two same digits
no of ways we can select 1 st one = 10 , but it must not be 0 = 9
no of ways we can select the 2nd one =1

step 3
no of ways we can select the last digit = 8

therefore ,
3*9*1*8=216
Re: How many 3 digit numbers can we make such that two of the digits are &nbs [#permalink] 30 Jul 2018, 07:04
Display posts from previous: Sort by

How many 3 digit numbers can we make such that two of the digits are

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.