Mechanics question- block on a wedge

I was wondering if anyone could offer some help with the following question:

A wedge with mass M rests on a frictionless horizontal table top. A block with mass m is placed on the wedge. There is no friction between the block and the wedge. The system is released from rest.

Calculate the acceleration of the wedge.

Express your answer in terms of M, m,α and g.

The answer I should get is:

http://session.masteringphysics.com/...9%5E%7B2%7D%7D

However I keep getting the following:

__-mgsinαcosα__

M+mcosαsinα

Does anyone see where I am going wrong? My working is as follows.

I will denote the normal force to the block by n.

n=mgcosα

Therefore the horizontal component of force on the wedge due to the block is given by F=-mgcosαsinα. The combined downward force due to the ramp and block is Mg+mgcosαsinα

Therefore (M+mcosαsinα)a=-mgcosαsinα (By Newtons second law)

Hence a=__mgsinαcosα__

M+mcosαsinα

I have no idea where I am going wrong so any help would be appreciated.