# Cardinality of Sets

#### bergausstein

##### Active member
i have solved these problem just want to make sure I'm on the right track.

1. Say the football team F, the basketball team B, and the track team T, decide to form a varsity club V. how many members will V have if $n\left(F\right)\,=\,25,\,n\left(B\right)\,=\,12,\,n\left(T\right)\,=\,30$ and no person belongs to two teams?

Solution. $n\left(F\right)+n\left(B\right)+n\left(T\right)\,=\,67$ there's no possibility of overlap.

a. If in problem 1, $n\left(F\cap T\right)\,=\,6$ but there are no members of B who are in F or T then what is $n\left(V\right)$?

Solution. since there are six persons who belong to T and F i will subtract 6 from 67 which is 61.

b. If in problem 1.a $n\left(F\cap T\right)\,=\,6$, $n\left(T\cap B\right)\,=\,4$ and $n\left(F\cap B\right)\,=\,0$ then what is $n\left(V\right)$?

Solution. there are now 10 persons that belong to two teams i will subtract these number from 67 and I will have 57.

c. if in problem 1.b $n\left(F\cap T\right)\,=\,6$, $n\left(T\cap B\right)\,=\,4$ and $n\left(F\cap B\right)\,=\,3$, and there are 2 three letter men that is $n\left(F\cap B\cap T\right)\,=\,2$, then what is $n\left(V\right)$?

in part C i don't understand the part where it say "there are 2 three letter men".

but this is what i Tried. there are now 13 persons who belong to two teams and 2 persons who are member of the three teams. 67-15 = 54-2 = 52.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
I agree with 1, 1a and 1b.

c. if in problem 1.b $n\left(F\cap T\right)\,=\,6$, $n\left(T\cap B\right)\,=\,4$ and $n\left(F\cap B\right)\,=\,3$, and there are 2 three letter men that is $n\left(F\cap B\cap T\right)\,=\,2$, then what is $n\left(V\right)$?

in part C i don't understand the part where it say "there are 2 three letter men".
As the problem says, "there are 2 three letter men" means $n\left(F\cap B\cap T\right)=2$.

but this is what i Tried. there are now 13 persons who belong to two teams and 2 persons who are member of the three teams. 67-15 = 54-2 = 52.
According to the inclusion–exclusion principle,$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$Therefore, you should add 2 instead of subtracting it.

#### bergausstein

##### Active member
does this ($n\left(F\cap B\cap T\right)$) mean that there are two persons who are member of the three teams? my thought process is like this: I don't want to count those 2 persons twice in the union that's why i subtracted it from the cardinality of union the three finite sets. like what i did with the 13 persons who are members of 2 two teams. please enlighten me further. thanks.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
does this ($n\left(F\cap B\cap T\right)$) mean that there are two persons who are member of the three teams?
No, $n(F\cap B\cap T)=2$ means that.

my thought process is like this: I don't want to count those 2 persons twice in the union that's why i subtracted it from the cardinality of union the three finite sets. like what i did with the 13 persons who are members of 2 two teams.
When you added $|F|+|B|+|T|$, you counted elements of $F\cap B\cap T$ three times. Then, when you subtracted $|F\cap B|$, $|F\cap T|$ and $|B\cap T|$, you subtracted elements of $F\cap B\cap T$ also three times because those 2 elements occur in each of the three intersections. As a result, they still need to be counted once.

#### bergausstein

##### Active member
thanks you're such a help!