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

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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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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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.

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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total digits = 6
6*4*4*3
288
IMO C

Total - restrictions
6P4-4P2 * 3! = 360-72=288

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

Thanks in advance.

Bunuel could you break this down please? Thank you

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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Yes, it did!
Thank you very much.

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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A simple process is that from the total number of 4 digits numbers that we can form using 1,2,5,6,8 and 9 without repetition, we subtract the cases where 2 and 5 are together, we get the cases where 2 and 5 are not together.

Calculating total number of 4 digit numbers we can form using 1,2,5,6,8 and 9 without

repetition;

This can be done by selecting any four digits among 1,2,5,6,8 and 9 in 6C4 ways and then arranging these four digits in 4! ways.

=6C4×4!

=(6!4!)/(2!×4!)

=6×5×4×3=360

Calculating total number of 4 digit numbers we can form using 1,2,5,6,8 and 9 without repetition where 2 and 5 are together;

Since, 2 and 5 have to be together, thus for the other two spaces the two digits from the rest of the 4 digits can be selected in 4C2 and these two selected digits along with 2 and 5 clubbed together can be arranged in 3! × 2! ways,

Thus, total ways in which 2 and 5 will be together = 4C2×3!×2!

= 4!2!×2!×3!×2!

=72

Thus, total ways in which 2 and 5 will not be together,

= 360 – 72

= 288

Hence, the correct answer is C.

We can choose 2 and 5 as adjacent and subtract from the total
6*5*4*3 = total number of arrangement or 360.
2 and 5 can sit together in
2, 5, number x, number y
Number x, 2, 5, number y
Number x, number y, 2, 5
So total of 3 options. So 3*
2/5 can be picked in 2 ways (5,2 or 2,5) so *2*
The other 2 numbers can be picked in 4*3 ways so *12
3*2*12=72
72 ways to sit 2and 5 together.
360-72 for not sitting them together.

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