1. ## [SOLVED] permutations

1. Of the 190 students graduating from Statsville high school, 111 played on a sports team, 26 served on the student council, and 67 were members of clubs. Only 10 students did not participate in any of these activities. If 13 students were members of both a team and a club, 9 were on both a team and the student council, and 11 were a member of both the student council and a club, how many graduating students have been a member of a team as well as the student council and a club. Use the principle of inclusion and exclusion to solve.

2. What have done towards working this problem?
You need to use a Venn diagram.
Please show what you can do with this.

3. Originally Posted by Plato
What have done towards working this problem?
You need to use a Venn diagram.
Please show what you can do with this.
Is there a way to do it without using a Venn Diagram?

4. Originally Posted by william
Is there a way to do it without using a Venn Diagram?
$\displaystyle 190 - 10 = \left| {S \cup C \cup M} \right| = \left| S \right| + \left| C \right| + \left| M \right| - \left| {S \cap C} \right| - \left| {S \cap M} \right| - \left| {C \cap M} \right| + \left| {S \cap C \cap M} \right|$

5. Hello, William!

By the way, this problem does not involve Permutations.

1. Of the 190 students graduating from Statsville high school:

111 played on a sports team
26 served on the student council
67 were members of clubs
. . 13 students were members of both a team and a club
. . 9 were on both a team and the student council
. . 11 were a member of both the student council and a club
. . . . 10 students did not participate in any of these activities.

How many graduating students were in all three activities?
There are formulas for this situation . . .

With two groups:
. . $\displaystyle n(A \cup B) \;=\;n(A) + n(B) - n(A \cap B)$

With three groups:
. . $\displaystyle n(A\:\cup\:B\:\cup C) \;=\;n(A) + n(B) + n(C) - n(A\,\cap\,B) - n(B\,\cap\,C)$ $\displaystyle - n(A\,\cap\,C) + n(A\,\cap\,B\,\cap C)$

Let:. . $\displaystyle \begin{array}{ccc}T &=& \text{played on a sport Team} \\ S &=& \text{served on Student council} \\ C &=& \text{a member of a Club}\end{array}$

There were: $\displaystyle 190 - 10 = 180$ students who were in at least one activity.
. . That is: .$\displaystyle n(T \cup S \cup C) \,=\,180$

So we have:

$\displaystyle \underbrace{n(T \,\cup\,S\,\cup C)}_{180} \:=\;\underbrace{n(T)}_{111} + \underbrace{n(S)}_{26} + \underbrace{n(C)}_{67} - \underbrace{n(T\,\cap\,S)}_{9} - \underbrace{n(S\,\cap\,C)}_{11}$ $\displaystyle - \underbrace{n(T\,\cap\,C)}_{13} + \underbrace{n(T\,\cap\,S\,\cap C)}_{x}$

. . . . . . $\displaystyle 180 \;=\;204 - 33 + x$

Therefore: .$\displaystyle x \:=\:9$

Edit: Too fast for me, Plato!
.

6. ## Soln of the problem

As it is given that ther are 190 students in all 10
Out of which do not participate in any of the events.
i.e. there are only 190 -10 = 180 students who participated in some event
or we can say that the union of all the paricipants of some event is of count 180 students.

Now we Let S=sport team particiapnt
C = student council
Cb = club member
then
190 - 10 = 180 = |S U C U Cb |= |S| + |C| + |Cb| - |S U C| - |SU cb| - |C U Cb| + |S intersection C intersection Cb|
Que is
|S intersection C intersection Cb|=?
|S intersection C intersection Cb|= |S U C U Cb |-|S| - |C| - |Cb| + |S U C| + |SU cb| + |C U Cb|

or |S intersection C intersection Cb|=180 - 111 - 26 - 67 + 13 + 11 + 9 = 9
ans = 9

7. Originally Posted by william
1. Of the 190 students graduating from Statsville high school, 111 played on a sports team, 26 served on the student council, and 67 were members of clubs. Only 10 students did not participate in any of these activities. If 13 students were members of both a team and a club, 9 were on both a team and the student council, and 11 were a member of both the student council and a club, how many graduating students have been a member of a team as well as the student council and a club. Use the principle of inclusion and exclusion to solve.
I know getting to know how to use the Venn Diagrams can be a little tricky, but you should practice with these diagrams before you commit to figuring out such problems without them. It will be much less confusing in the long run.