In how many ways he can invite atleast one of his friends for dinner if a man has $7$ friends? 
A man has 7 friends.In how many ways he can invite atleast one of his friends for dinner?

Options
a)$63$   b)$120$   c)$127$   d)$256$
My approach:
@Edit
There are $2$ ways whether he invites or not invites.
I am thinking of $2^7-1$ ways because  $1$st friend can be invited in $7C1$ times  2nd friend can be invited in $7C1$ times etc and also only 1 condition where no one is invited have to be subtracted.

Is it right?

 A: Hint
The contrary is "in how many way can he invite no body ?"
Notice that if $\Omega$ represent all the way he can invite his friends, then $|\Omega|=2^7$. But actually, the question is not very well asked. 
A: Every friend corresponds to a binary decision variable, to invite or not invite. If we write a binary number (ordered sequence of binary digits), letting 0 mean "do not invite" and 1 mean "invite", then the number will be 7 binary digits long and therefore hold $2^7$ different numbers. However in order to have "at least one friend" invited, the number corresponding to inviting no one: 0000000 is "forbidden", leaving us with $2^7-1 = 127$ configurations left.
A: I advise that you read up on the choose function or Binomial coefficient ${n \choose k}$ , which counts the number of ways to pick a subset of $0\le k\le n$ objects from a set of $n$ distinguishable objects. In this case, you should think of the man's friends as distinguishable objects as they are not all alike. In other combinatorial problems, such as some ball and urn problems, the objects are indistinguishable and so such an approach is not appropriate.
See: https://en.wikipedia.org/wiki/Binomial_coefficient
