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

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

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New post Updated on: 27 May 2018, 04:27
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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.
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Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

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New post 28 May 2018, 06:39
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
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Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

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New post 27 May 2018, 05:30
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.
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How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

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New post 27 May 2018, 05:57
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.
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Re: How many 4 digit numbers can be formed using the digits 1, 2, 5, 6, 8  [#permalink]

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New post 06 Jun 2018, 22:11
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.
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Re: How many 4 digit numbers can be formed using the digits 1, 2, 5, 6, 8  [#permalink]

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New post 08 Jun 2018, 02:37
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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?
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Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

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New post 21 Jan 2019, 07:52
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
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Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

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New post 21 Jan 2019, 11:42
Total - restrictions
6P4-4P2 * 3! = 360-72=288
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How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

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New post 21 Jan 2019, 22:53
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 ??

Thanks in advance.
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Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

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New post 22 Jan 2019, 01:43
Bunuel could you break this down please? Thank you
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How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

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New post 25 Jan 2019, 23:35
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.
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Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

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New post 25 Jan 2019, 23:52
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.
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Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

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New post 27 Jan 2019, 19:16
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.
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Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and  [#permalink]

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New post 23 Aug 2019, 00:11
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) ?
Please clarify

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Re: How many 4 digit numbers can be formed using the digits 1,2,5,6,8 and   [#permalink] 23 Aug 2019, 00:11
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