let the numbers be i1,i2,i3....i8

Array A has to contain 5 sorted numbers..
So basically u have to make 5 choices among given 8. Standard formula of it
is 8C5 (8!/(3! * 5!)).

After creating array A there will be 3 elements left which will suffice to
create array B (no choice is left in this case so it will be 1C1=1).

No shuffling among the numbers in array A and B will be there as numbers
have to be sorted and so only one possible arrangment is possible to
maintain the array sorted.

so total number of arrangements will be 8C5 * 1C1 = 8C5= 56.



On Wed, Oct 13, 2010 at 5:25 PM, Divya Jain <[email protected]>wrote:

> ignore the above post of mine
> @ shiyam
> u r doing wrong. we hv been the size of both array b and c which is 5 n 3
> respectively
>
> @ AlgoSAu
> can u please explain ur method?
>
>
> On 13 October 2010 17:23, Divya Jain <[email protected]> wrote:
>
>> @above
>> can u plz explain hw did u apply this formula?
>>
>>
>> On 13 October 2010 16:14, Shiyam code_for_life 
>> <[email protected]>wrote:
>>
>>> Lets say 8 integers are 1 to 8
>>>
>>> So first way array can be split is 1,7
>>>
>>> 1st array can have one among 8 integers so in 8 ways and 2nd array has
>>> just one arrangement of other integers in sorted manner and same goes
>>> on for other ways to split the array.
>>>
>>> Whats wrong here?
>>>
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