Let $A=\{1,2,\cdots ,100\}$. Let $S$ be some set of subsets of $A$ such that any two elements of $S$ have a nonempty intersection. Then what is the maximum possible cardinality of $S$?
My answer- My intuition is that all the subsets must have one element in common, so if I consider the element the element $1$, then the sets {$1$},{$1$,any one among 99 other elements},{$1$, any one among 99 other elements, any one among 98 other elements} $,\cdots$ until the original set $A$. This can be done in ${99}\choose_0$ + ${99}\choose_{1}$ + $\cdots$ + $99\choose_{99}$ ways which is $2^{99}$. Please can anyone verify the way I have solved the problem. Thank you.