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

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.

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?

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.

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

Posted from my mobile device

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.

Posted from my mobile device

The key to solving the problem is getting the right combination
let us calculate the totall no ways of forming non repeating numbers=6*5*4*3 =360
total no of forming with 2 and 5 = 24*3
which arise from the possibility = 4*3ab or ab4*3 or 4ab*3
Therefore the total no of ways = 360-72= 288
Hence IMO C

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