dont multiply by 5! and 3! because sub-arrays have to remain sorted too so
there will be only ONE possible arrangement of 5 elements and 3 elements in
sub-array.

On Wed, Oct 13, 2010 at 9:27 AM, sudheer babu <[email protected]>wrote:

> 8C5*(5!*3!)=8!
>
>
> On Tue, Oct 12, 2010 at 9:31 PM, ANUJ KUMAR <[email protected]>wrote:
>
>> 56
>>
>> On Tue, Oct 12, 2010 at 5:09 PM, divya <[email protected]> wrote:
>> > Given a sorted array A with 8 integers.We want to have 2 sorted arrays
>> > B(with 5 integers)
>> > And C(with 3 integers) such that merging B and C would result in A.
>> > How many such pairs exists
>> >
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