santro789 wrote:
Can someone explain this with any other method?
If it requires to have the sum of 2
Then arrangement will be (1 1 0)
{(1/2)(1/2)(1/2)}(3!/2!)
=(1/8)*3
Can someone explain this with any other method?
If it requires to have the sum of 2
Then arrangement will be (1 1 0)
{(1/2)(1/2)(1/2)}(3!/2!)
=(1/8)*3