A probability problem by Mohammad Khaza

In the problem, the number of friends he invite at a time was not specified. Therefore, he may invite only one, or only two, or only three and so on. The number of ways is given by the formula: . $C=2^n-1$

Substituting, we get

$C=2^6-1=64-1=$ $\boxed{63}$

Let's see the inviting from one friend's point of view. He/she is invited or he/she isn't invited to the party, that mean $2$ possibilities. So for each friend there are $2$ possibilities, which means there are $2*2*2*2*2*2=2^6=64$ ways of inviting. But be careful, the person can't invite $0$ friends, so the answer is $64-1=\boxed{63}$ .

The combinations if $k$ friends invited from 6 is given by ${6 \choose k}$ , where $k = 1,2,3,4,5,6$ . Then the total number of combinations possible is:

$\begin{aligned} N & = \sum_{\color{#3D99F6}k=1}^6 {6 \choose k} = \sum_{\color{#D61F06}k=0}^6 {6 \choose k} - 1 = 2^6 - 1 = \boxed{63} \end{aligned}$

this will be : nCr = 6C1 +6C2+6C3 =6C4+6C5+6C6

                          =    6+15+20+15+6+1 

                          = 63

there is 63 ways.