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

It is currently 13 Oct 2019, 21:10

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

Sid intended to type a seven-digit number, but the two 3's he meant to

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

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58335
Sid intended to type a seven-digit number, but the two 3's he meant to  [#permalink]

Show Tags

New post 01 Apr 2015, 04:59
4
13
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

68% (01:54) correct 32% (02:11) wrong based on 152 sessions

HideShow timer Statistics

Director
Director
User avatar
Joined: 07 Aug 2011
Posts: 502
Concentration: International Business, Technology
GMAT 1: 630 Q49 V27
GMAT ToolKit User
Re: Sid intended to type a seven-digit number, but the two 3's he meant to  [#permalink]

Show Tags

New post 01 Apr 2015, 05:36
1
Bunuel wrote:
Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead was the five-digit number 52115. How many different seven-digit numbers could Sid have meant to type?

A. 10
B. 16
C. 21
D. 24
E. 27


Kudos for a correct solution.


Should be 21.

there are two possibilities for placing 2 3s .

case 1: two 3s were missed consecutively. i.e. he typed 33 and it came blank on screen.
-5-2-1-1-5- in this arrangement we can fit 33 in 6 ways . (Six dashes, each dash represent one possible place for placing 33)

case 2: two 3s are not together, i.e. they have one or more digits between them .
-5-2-1-1-5- , in this arrangement
if we place first 3 at first dash i.e. 35-2-1-1-5- then the other 3 can fit into 5 places.
if we place first 3 at second dash i.e. -532-1-1-5- then the other 3 can fit into 4 places.
if we place first 3 at third dash i.e. -5-231-1-5- then the other 3 can fit into 3 places.
if we place first 3 at fourth dash i.e. -5-2-131-5- then the other 3 can fit into 2 places.
if we place first 3 at Fifth dash i.e. -5-2-1-135- then the other 3 can fit into 1 place.
so total 15 ways.

case 2 + case 1 = 6+ 15 = 21 ways

Answer C
Retired Moderator
avatar
B
Status: On a mountain of skulls, in the castle of pain, I sit on a throne of blood.
Joined: 30 Jul 2013
Posts: 300
GMAT ToolKit User Reviews Badge
Re: Sid intended to type a seven-digit number, but the two 3's he meant to  [#permalink]

Show Tags

New post 01 Apr 2015, 05:45
Bunuel wrote:
Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead was the five-digit number 52115. How many different seven-digit numbers could Sid have meant to type?

A. 10
B. 16
C. 21
D. 24
E. 27


Kudos for a correct solution.


The other numbers are fixed so we have to choose 2 spaces for the 2 numbers.

7C2=21


Answer : C
Manager
Manager
avatar
Joined: 18 Sep 2014
Posts: 226
Reviews Badge
Re: Sid intended to type a seven-digit number, but the two 3's he meant to  [#permalink]

Show Tags

New post 01 Apr 2015, 06:00
1
Bunuel wrote:
Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead was the five-digit number 52115. How many different seven-digit numbers could Sid have meant to type?

A. 10
B. 16
C. 21
D. 24
E. 27


Kudos for a correct solution.


I just interpreted the question in a simple fashion : How to place 2 in a 7 digit number and how many ways can we do it?

Simple 7C2 = 21 ways.
Hence C
_________________
Kindly press the Kudos to appreciate my post !! :-)
Manager
Manager
avatar
Joined: 17 Mar 2015
Posts: 115
Sid intended to type a seven-digit number, but the two 3's he meant to  [#permalink]

Show Tags

New post 02 Apr 2015, 02:42
2
My idea was as follows, I'm not exactly great in these kind of questions so it would be nice if someone commented.
We got 2 numbers to place.
1st one can be placed in one of the 6 spots: _ X _ X _ X _ X _ X _ (6 spots = 6 underscores), we can do it in 6 different ways
2nd number can be placed in one of the 7 spots (after having placed the first number we got a 6 digit now: _ X _ X _ X _ X _ X _ X _ : 7 spots = 7 underscores), respectively 7 ways.
As a result we get 6*7 = 42 ways
Since we got duplicates coz of numbers being equal, we'll have to cut it in half, which gives us 42/2 = 21, the answer C.
Manager
Manager
avatar
Joined: 18 Dec 2014
Posts: 99
Re: Sid intended to type a seven-digit number, but the two 3's he meant to  [#permalink]

Show Tags

New post 02 Apr 2015, 02:49
1
7 places, 2 choices.

7! / 2!5!

7*6*5*4*3*2*1 / 2*1*5*4*3*2*1 = 7*6 / 2*1 = 21.
Manager
Manager
avatar
B
Joined: 26 Dec 2012
Posts: 144
Location: United States
Concentration: Technology, Social Entrepreneurship
WE: Information Technology (Computer Software)
Re: Sid intended to type a seven-digit number, but the two 3's he meant to  [#permalink]

Show Tags

New post 04 Apr 2015, 20:10
We can choose ways to place two 3 three in a 7 digit integer for which 5 digits are fixed is by 7C2 ways.
7C2 = 21
Hence answer is C
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 7949
Sid intended to type a seven-digit number, but the two 3's he meant to  [#permalink]

Show Tags

New post 04 Apr 2015, 21:17
1
1
Bunuel wrote:
Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead was the five-digit number 52115. How many different seven-digit numbers could Sid have meant to type?

A. 10
B. 16
C. 21
D. 24
E. 27


Kudos for a correct solution.


two ways of fixing two 3's..
both together- any six places-6 ways..
both separately- any two places out of available six places- 6C2=15..
total 21 ways.. C

7C2 will not stand for all Q of these types..
example if instead of two 3's , say it were 3 and 7.. the ans would be 7P2, as order would matter...
_________________
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58335
Re: Sid intended to type a seven-digit number, but the two 3's he meant to  [#permalink]

Show Tags

New post 06 Apr 2015, 06:09
1
1
Bunuel wrote:
Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead was the five-digit number 52115. How many different seven-digit numbers could Sid have meant to type?

A. 10
B. 16
C. 21
D. 24
E. 27


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

First of all, notice that the numerals 5-2-1-1-5 must be in that order --- the order may be broken at any point but one or two 3's, but ignoring the 3's, those five must have that order.

As is often the case, in this counting problem we have a variety of ways to frame the question. I believe the most straightforward is as follows. Consider the seven "blank spaces" into which we will write the seven digits of the original number.

_ _ _ _ _ _ _

Now, select any two of those spaces, and put the two 3's there. The two 3's could be next to each other or apart. Suppose, for example, we put them here:

_ _ 3 _ _3 _

At this point, notice that everything else about the number is determined: we simply will put the digits 5-2-1-1-2 in order in the remaining spaces. Just picking two of the seven spaces for the location of the two 3's is enough to determine the entire original number. Well, there are 7C2 ways of selecting two slots from seven, so 7C2 must be the number of possible original numbers that Sid intended to write.

See this post for various techniques on calculating combinations.

7C2 = 21

So there are 21 possible original numbers that Sid could have intended.

Answer = C
_________________
VP
VP
avatar
P
Joined: 12 Dec 2016
Posts: 1493
Location: United States
GMAT 1: 700 Q49 V33
GPA: 3.64
GMAT ToolKit User
Re: Sid intended to type a seven-digit number, but the two 3's he meant to  [#permalink]

Show Tags

New post 27 Nov 2017, 13:59
the trap here is that it is not 6C2, but 7C2 b/c 2 numbers of 3 can stand next to each other.
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 13081
Re: Sid intended to type a seven-digit number, but the two 3's he meant to  [#permalink]

Show Tags

New post 09 Feb 2019, 23:41
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Bot
Re: Sid intended to type a seven-digit number, but the two 3's he meant to   [#permalink] 09 Feb 2019, 23:41
Display posts from previous: Sort by

Sid intended to type a seven-digit number, but the two 3's he meant to

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





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