Intersecting family 
Let $X = \{1,...,7\} \ $ and let $B$ consist of the seven subsets $$\{\{1,2,3\}, \{1,4,5\}, \{1,6,7\},\{2,4,6\},\{2,5,7\},\{3,4,7\},\{3,5,6\}\}.$$ Let $F$ be the set of all those subsets of $X$ which contain a member of $B$. What is $|F|$?

Since B is an intersecting family, we therefore have that $B\subset F$ and thus I can use a proposition that states, an intersecting family of subsets of $\{1,...,n\} \ $ satisfies $|F|\le 2^{n-1}$ and thus there are $2^{n-1}$ intersecting families, but how can I go about solving this without using that given proposition? Because I can't see why $|F|$ is $2^{7-1} = 64$.
 A: The family $F$ contains :


*

*Subsets of cardinal $3$, there is exactly $7$ elements in $F$ with this propriety which are the elements of $B$

*Subsets of cardinal $4$, How can we construct elements elements of $F$ of cardinal $4$? by adding to each subset of $B$ another element, e.g $\{1,2,3\}$ one can add $4,5,6,7$ to this set and construct the $4$ elements:$\{1,2,3,4\},\{1,2,3,5\},\{1,2,3,6\},\{1,2,3,7\}\in F$, and the same thing can be done for others in total there $4\cdot 7=28$ elements of cardinal $4$ in $F$.

*Subsets of cardinal $5$, there is exactly ${7 \choose 5}$ subset of cardinal $5$, and every subset of cardinal $5$ contain an element of the set $B$ hence every subset of cardinal $5$ is an element of $F$, so there is ${7 \choose 5}$ of elements of $F$ of cardinal $5$.

*Subsets of cardinal $6$, there is exactly ${7 \choose 6}$ subset of cardinal $5$ and all of them are elements of $F$ because each subset of cardinal $6$ contains an element of $B$

*One subset of cardinal $7$ which is in $F$


Hence we have $|F|=7+28+21+7+1=64$
A: You want to break up $\Bbb P(X)$ into complementary pairs $\{A, A^c\}$.  Since $B$ is an intersecting family, you know that at most one element of each pair lies in $F$.  Since you want to show that $|F| = 64 = |\Bbb P(X)|/2$, you need an argument that for every complementary pair, one of the two lies in $F$.  The most elegant way I can think to prove this is relies on the following:


*

*Claim 1: If $A$ is a set with 4 elements not lying in $F$, then $A^c$ is an element of $B$.

*Claim 2: Every subset of $X$ with at least 5 elements lies in $F$.


To prove the first claim, we count the number of $4$-element sets lying in $F$.  Note that each sets contain a unique element of $B$, since the intersection of two elements of $B$ is only one element.   For every element of $B$, there are four elements you can add.  Therefore, there are $|B|*4 = 7$ sets with four elements in $F$.  These plus the 7 complements of the elements of $B$ account for all such subsets.
To prove the second claim, consider the seven elements $C$ that are complements of elements of $B$:
$$
C := \{\{ 4,5, 6, 7\},
\{2, 3, 6, 7\},
\{2, 3, 4, 5\},
\{1, 3, 5, 7\},
\{1, 3, 4, 6\},
\{1, 2, 5, 6\},
\{1, 2, 4, 7\}\}
$$
It's easy to see that since any two elements of $B$ intersect each other in one point, that any two elements of $C$ contain precisely two elements in common.
Now, any five element subset $\{a,b,c,d,e\}$ of $X$ not lying in $F$ must have each of its four-element subsets in $C$, by the first claim.  However, any two such subsets, for instance $\{a,b,c,d\}$ and $\{a,b,c,e\}$, have more than two elements in common.  This contradiction establishes the second claim.
