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

 It is currently 14 Oct 2019, 19:42 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.  How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and

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

Hide Tags

Intern  B
Joined: 20 Apr 2018
Posts: 9
How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

Show Tags

2
12 00:00

Difficulty:   65% (hard)

Question Stats: 57% (02:14) correct 43% (02:14) wrong based on 107 sessions

HideShow timer Statistics

How many 4 digit numbers can be formed using the digits 1, 2, 5, 6, 8 and 9 without repetition, so that 2 and 5 never come together ?

A. 120
B. 180
C. 288
D. 360
E. 720

Originally posted by failatmath on 27 May 2018, 04:21.
Last edited by Bunuel on 27 May 2018, 04:27, edited 1 time in total.
Renamed the topic and edited the question.
Manager  B
Joined: 08 Sep 2016
Posts: 102
Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

Show Tags

5
1
Total numbers = 6 . (1,2,5,6,8,9)

Total ways = 6*5*4*3 = 360 . (any 6 numbers can take the first slot, any 5 in the second slot ...)

Restriction - 2 and 5 can't be placed right next to each other.

X= 2
Y= 5

First scenario: XY43 = 1*1*4*3 = 12 . Multiply it by 2 since X and Y can switch positions too. = 24
Second scenario: 4XY3 = 4*1*1*3 = 12 Multiply it by 2 since X and Y can switch positions too. = 24
Third scenario: 43XY = 4*3*1*1 = 12 . Multiply it by 2 since X and Y can switch positions too. = 24

Total - unwanted = 360 -24 -24 - 24 = 288

Option C
General Discussion
examPAL Representative P
Joined: 07 Dec 2017
Posts: 1140
Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

Show Tags

1
failatmath wrote:
How many 4 digit numbers can be formed using the digits 1, 2, 5, 6, 8 and 9 without repetition, so that 2 and 5 never come together ?

A. 120
B. 180
C. 288
D. 360
E. 720

Problems that ask us to arrange objects under specific constraints can often be split into smaller, simpler problems.
We'll look for such a division, a Logical approach.

We'll split into cases:

If we form a number without 2 and without 5, then we must use the numbers 1,6,8,9.
This gives 4! = 24 options.

If we pick only one of 2,5 and another 3 of the remaining numbers this gives 2C1 * 4C3 = 8 options.
This gives 8 * 4! = 192 arrangements (all of which are good as 2,5 are not together)

If we pick 2, 5, and another 2 of the remaining numbers this gives 2C2 * 4C2 = 6 options.
How many 'bad' options, that is options such that 2 and 5 are together, are there?
3! * 2: 2-5-x-y, 2-5-y-x, x-2-5-y, y-2-5-x, x-y-2-5, y-x-2-5 and all of these with the numbers 2,5 reversed.
This is essentially arranging 3 objects: x,y, and a '2-5' clump.
So - we have (4! - 3!*2) = 12 arrangements per option times 6 options giving 72 arrangements.

24 + 192 + 72 = 288

(C) is our answer.
_________________
Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3333
Location: India
GPA: 3.12
How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

Show Tags

3
failatmath wrote:
How many 4 digit numbers can be formed using the digits 1, 2, 5, 6, 8 and 9 without repetition, so that 2 and 5 never come together ?

A. 120
B. 180
C. 288
D. 360
E. 720

Since there are 6 digits and we have to form a 4-digit number,
there are $$C_4^6 = 15$$ possibilities. The 4 digits can be arranged in $$4!$$ or $$24$$ ways.

The total number of 4 digit numbers possible without repetition are $$15*24 = 360$$.

The remaining two digits of the 4-digit number(where digits 2 and 5 are together)
can be found in $$C_2^4 = 6$$ ways. There are $$3!*2(12)$$ ways of arranging these 4 digits.
We multiply the total number of arrangements by 2 is because 25 is different from 52.

The total number of 4 digit numbers possible such that 2 and 5 are together are $$6*6*2 = 72$$.

Therefore, there are 360 - 72 = 288(Option C) numbers possible such that 2 and 5 don't come together.
_________________
You've got what it takes, but it will take everything you've got
Intern  Joined: 08 May 2018
Posts: 5
Re: How many 4 digit numbers can be formed using the digits 1, 2, 5, 6, 8  [#permalink]

Show Tags

There are 6 digits in total and we have to form a 4 digit number, hence that can be done in 6p4 ways.
The total number of 4 digit numbers possible without repetition are 6!/2!=360.

Numbers where 6 and one another number are together can be found out as 4c2 (because out of 6 digits ,2 are always together,remaining digits =4)

4c2=6 ways
Also digits can themselves be arranged in 3! ways =6.
And 6 and other number can be arranged amongst themselves,so total nuber of ways become 2(6*6)=72.

hence total ways =360-72=288
c is the answer.
Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3333
Location: India
GPA: 3.12
Re: How many 4 digit numbers can be formed using the digits 1, 2, 5, 6, 8  [#permalink]

Show Tags

1
1
1
GMATSkilled wrote:
How many 4 digit numbers can be formed using the digits 1, 2, 5, 6, 8 & 9, without repetition, so that 2 & % are not together?

A. 120

B. 180

C. 288

D. 360

E. 720

We need to form a 4 digit number(with 6 digits available). There are $$C_4^6 = 15$$ ways of choosing the digits.

There are 4! or 24 ways of arranging these digits. The 4 digit numbers possible are $$24*15 = 360$$.

Possibilities of numbers with 2 and 5 together

There are $$C_2^4 = 6$$ ways of choosing the other 2 digits and $$3! = 6$$ ways of arranging the digits.
2 and 5 can be arranged ways: ways : as 25 and as 52. The 4 digit numbers possible are $$2*6*6 = 72$$.

Therefore, the 4 digit numbers possible when the digits 2 and 5 are not together are 360 - 72 = 288(Option C)

GMATSkilled - I believe there is a typo in this question. Can you please edit it?
_________________
You've got what it takes, but it will take everything you've got
GMAT Club Legend  D
Joined: 18 Aug 2017
Posts: 4987
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

Show Tags

1
failatmath wrote:
How many 4 digit numbers can be formed using the digits 1, 2, 5, 6, 8 and 9 without repetition, so that 2 and 5 never come together ?

A. 120
B. 180
C. 288
D. 360
E. 720

total digits = 6
6*4*4*3
288
IMO C
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
Intern  B
Joined: 27 Nov 2018
Posts: 30
Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

Show Tags

Total - restrictions
6P4-4P2 * 3! = 360-72=288
Manager  B
Joined: 17 Jun 2017
Posts: 60
How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

Show Tags

pushpitkc wrote:
GMATSkilled wrote:
How many 4 digit numbers can be formed using the digits 1, 2, 5, 6, 8 & 9, without repetition, so that 2 & % are not together?

A. 120

B. 180

C. 288

D. 360

E. 720

We need to form a 4 digit number(with 6 digits available). There are $$C_4^6 = 15$$ ways of choosing the digits.

There are 4! or 24 ways of arranging these digits. The 4 digit numbers possible are $$24*15 = 360$$.

Possibilities of numbers with 2 and 5 together

There are $$C_2^4 = 6$$ ways of choosing the other 2 digits and $$3! = 6$$ ways of arranging the digits.
2 and 5 can be arranged ways: ways : as 25 and as 52. The 4 digit numbers possible are $$2*6*6 = 72$$.

Therefore, the 4 digit numbers possible when the digits 2 and 5 are not together are 360 - 72 = 288(Option C)

GMATSkilled - I believe there is a typo in this question. Can you please edit it?

Can you please explain this step 3!=6 ways of arranging the digits. ? why did we consider 3! again ??

Manager  B
Joined: 28 Jun 2018
Posts: 73
Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

Show Tags

Bunuel could you break this down please? Thank you
Intern  B
Joined: 04 Oct 2016
Posts: 17
How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

Show Tags

All answers are good. But, I am unable to find what is wrong with the below method of finding the number of ways 2 and 5 are together.

- - - -
4 3 2 5

Since, 2 and 5 are fixed in a position, rest 2 positions (6-2) can be 4*3 =12. And the 2 numbers (2 and 5) can be arranged among themselves in 2! ways
=> 12*2=24

Please advise what else am I missing here.
Director  G
Joined: 09 Mar 2018
Posts: 997
Location: India
Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

Show Tags

mischiefmanaged wrote:
All answers are good. But, I am unable to find what is wrong with the below method of finding the number of ways 2 and 5 are together.

- - - -
4 3 2 5

Since, 2 and 5 are fixed in a position, rest 2 positions (6-2) can be 4*3 =12. And the 2 numbers (2 and 5) can be arranged among themselves in 2! ways
=> 12*2=24

Please advise what else am I missing here.

Even i solved the same way,

Inline is the approach

25 when the are together, it should be 12*3= 36 and not 12*2

2 5 - - , - 2 5 - , - - 25, not these can be filled in 36 way

2 5 can be 5 2, 36 ways will become 72 ways

360 - 72 ways will give you 288 ways

Let me know if this helps you or not.
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Intern  B
Joined: 04 Oct 2016
Posts: 17
Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

Show Tags

KanishkM wrote:
mischiefmanaged wrote:
All answers are good. But, I am unable to find what is wrong with the below method of finding the number of ways 2 and 5 are together.

- - - -
4 3 2 5

Since, 2 and 5 are fixed in a position, rest 2 positions (6-2) can be 4*3 =12. And the 2 numbers (2 and 5) can be arranged among themselves in 2! ways
=> 12*2=24

Please advise what else am I missing here.

Even i solved the same way,

Inline is the approach

25 when the are together, it should be 12*3= 36 and not 12*2

2 5 - - , - 2 5 - , - - 25, not these can be filled in 36 way

2 5 can be 5 2, 36 ways will become 72 ways

360 - 72 ways will give you 288 ways

Let me know if this helps you or not.

Yes, it did!
Thank you very much.
Intern  B
Joined: 11 May 2019
Posts: 19
Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

Show Tags

Funsho84 So had it been how many 6 digit can be formed using those numbers would have been
(6×5×4×3×2×1) - (5×4×3×2×1 ×2) ?

Posted from my mobile device Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and   [#permalink] 23 Aug 2019, 00:11
Display posts from previous: Sort by

How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and

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

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