Question

when the members of a family discussed where their annual reunion should take place, they found that out of all the family members, 10 would not go to a park, 6 would not go to a beach, 9 would not go to the family cottage, 3 would go to neither a park nor a beach, 2 would go to neither a beach nor the family cottage, 6 would go to neither a park nor the family cottage, 1 would not go to a park or a beach or to the family cottage, and 1 would go to all three places. what is the total number of family members? there are \boxed{} total family members. (simplify your answer.)

Explanation:

Let \( P \) be the set of those who go to the park, \( B \) to the beach, and \( C \) to the cottage. Let \( n(X) \) denote the number of elements in set \( X \), and \( n(\overline{X}) \) the complement. We know:
- \( n(\overline{P}) = 10 \), so \( n(P) = N - 10 \) (where \( N \) is total members).
- \( n(\overline{B}) = 6 \), so \( n(B) = N - 6 \).
- \( n(\overline{C}) = 9 \), so \( n(C) = N - 9 \).
- \( n(\overline{P} \cap \overline{B}) = 3 \), so \( n(P \cup B) = N - 3 \). By inclusion - exclusion, \( n(P) + n(B)-n(P \cap B)=N - 3 \), so \( n(P \cap B)=(N - 10)+(N - 6)-(N - 3)=N - 13 \).
- \( n(\overline{B} \cap \overline{C}) = 2 \), so \( n(B \cup C)=N - 2 \). Then \( n(B)+n(C)-n(B \cap C)=N - 2 \), so \( n(B \cap C)=(N - 6)+(N - 9)-(N - 2)=N - 13 \).
- \( n(\overline{P} \cap \overline{C}) = 6 \), so \( n(P \cup C)=N - 6 \). Then \( n(P)+n(C)-n(P \cap C)=N - 6 \), so \( n(P \cap C)=(N - 10)+(N - 9)-(N - 6)=N - 13 \).
- \( n(\overline{P} \cap \overline{B} \cap \overline{C}) = 1 \), so \( n(P \cup B \cup C)=N - 1 \).
- \( n(P \cap B \cap C)=1 \).

By the principle of inclusion - exclusion for three sets:
\( n(P \cup B \cup C)=n(P)+n(B)+n(C)-n(P \cap B)-n(P \cap C)-n(B \cap C)+n(P \cap B \cap C) \)

Substitute the expressions:
\( N - 1=(N - 10)+(N - 6)+(N - 9)-(N - 13)-(N - 13)-(N - 13)+1 \)

Step 1: Simplify the right - hand side

First, expand the right - hand side:
\( (N - 10)+(N - 6)+(N - 9)-(N - 13)-(N - 13)-(N - 13)+1 \)
\( = 3N-(10 + 6+9)-3N + 3\times13 + 1 \)
\( = 3N - 25-3N+39 + 1 \)
\( = 15 \)

Step 2: Solve for \( N \)

We have the equation \( N - 1=15 \)
Add 1 to both sides: \( N=15 + 1=16 \)? Wait, no, let's re - do the expansion.

Wait, let's do the expansion term by term:

\( (N - 10)+(N - 6)+(N - 9)=3N-25 \)

\( -(N - 13)-(N - 13)-(N - 13)=-3N + 39 \)

Then \( 3N-25-3N + 39+1=15 \)

So \( N - 1 = 15\)? No, wait, the left - hand side is \( n(P\cup B\cup C)=N - 1 \)

Wait, let's use another approach. Let's define the regions:

Let \( a=n(P\cap B\cap C)=1 \)

Let \( x=n(P\cap B\cap\overline{C}) \), \( y=n(P\cap\overline{B}\cap C) \), \( z=n(\overline{P}\cap B\cap C) \)

We know that \( n(\overline{P}\cap\overline{B}) = 3 \), which includes \( n(\overline{P}\cap\overline{B}\cap C)+n(\overline{P}\cap\overline{B}\cap\overline{C}) \). Since \( n(\overline{P}\cap\overline{B}\cap\overline{C}) = 1 \), then \( n(\overline{P}\cap\overline{B}\cap C)=3 - 1 = 2 \), so \( z = 2 \)

\( n(\overline{B}\cap\overline{C}) = 2 \), which includes \( n(\overline{P}\cap\overline{B}\cap\overline{C})+n(P\cap\overline{B}\cap\overline{C}) \). So \( n(P\cap\overline{B}\cap\overline{C})=2 - 1 = 1 \)

\( n(\overline{P}\cap\overline{C}) = 6 \), which includes \( n(\overline{P}\cap\overline{B}\cap\overline{C})+n(\overline{P}\cap B\cap\overline{C}) \). So \( n(\overline{P}\cap B\cap\overline{C})=6 - 1 = 5 \)

Now, \( n(\overline{P}) = 10 \), which is \( n(\overline{P}\cap\overline{B}\cap\overline{C})+n(\overline{P}\cap B\cap\overline{C})+n(\overline{P}\cap\overline{B}\cap C)+n(\overline{P}\cap B\cap C)=1 + 5+2 + z \)? Wait, no, \( n(\overline{P})=n(\overline{P}\cap\overline{B}\cap\overline{C})+n(\overline{P}\cap B\cap\overline{C})+n(\overline{P}\cap\overline{B}\cap C)+n(\overline{P}\cap B\cap C) \)

We know \( n(\overline{P}\cap\overline{B}\cap\overline{C}) = 1 \), \( n(\overline{P}\cap B\cap\overline{C}) = 5 \), \( n(\overline{P}\cap\overline{B}\cap C)=2 \), and \( n(\overline{P}\cap B\cap C)=z \)

So \( 1+5 + 2+z=10\Rightarrow z = 2 \) (which we already found)

\( n(\overline{B}) = 6 \), which is \( n(\overline{P}\cap\overline{B}\cap\overline{C})+n(P\cap\overline{B}\cap\overline{C})+n(\overline{P}\cap\overline{B}\cap C)+n(P\cap\overline{B}\cap C) \)

\( 1 + 1+2+x=6\Rightarrow x = 2 \)

\( n(\overline{C}) = 9 \), which is \( n(\overline{P}\cap\overline{B}\cap\overline{C})+n(P\cap\overline{B}\cap\overline{C})+n(\overline{P}\cap B\cap\overline{C})+n(P\cap B\cap\overline{C}) \)

\( 1+1 + 5+y=9\Rightarrow y = 2 \)

Now, the total number of family members is the sum of all regions:

\( n(\overline{P}\cap\overline{B}\cap\overline{C})+n(\overline{P}\cap B\cap\overline{C})+n(P\cap\overline{B}\cap\overline{C})+n(\overline{P}\cap\overline{B}\cap C)+n(P\cap B\cap\overline{C})+n(P\cap\overline{B}\cap C)+n(\overline{P}\cap B\cap C)+n(P\cap B\cap C) \)

\( =1 + 5+1+2 + 2+2+2 + 1=16 \)? Wait, no, let's list all regions:

- Region 1: \( \overline{P}\cap\overline{B}\cap\overline{C} \): 1
- Region 2: \( \overline{P}\cap B\cap\overline{C} \): 5
- Region 3: \( P\cap\overline{B}\cap\overline{C} \): 1
- Region 4: \( \overline{P}\cap\overline{B}\cap C \): 2
- Region 5: \( P\cap B\cap\overline{C} \): 2 (x)
- Region 6: \( P\cap\overline{B}\cap C \): 2 (y)
- Region 7: \( \overline{P}\cap B\cap C \): 2 (z)
- Region 8: \( P\cap B\cap C \): 1 (a)

Now sum them up: \( 1+5 + 1+2+2+2+2+1=16 \)? Wait, but let's check with the inclusion - exclusion formula again.

Let \( N \) be the total number of family members.

We know that:

\( n(\overline{P})=10\Rightarrow n(P)=N - 10 \)

\( n(\overline{B})=6\Rightarrow n(B)=N - 6 \)

\( n(\overline{C})=9\Rightarrow n(C)=N - 9 \)

\( n(\overline{P}\cap\overline{B}) = 3\Rightarrow n(P\cup B)=N - 3 \). By \( n(P\cup B)=n(P)+n(B)-n(P\cap B) \), we have \( n(P\cap B)=(N - 10)+(N - 6)-(N - 3)=N - 13 \)

\( n(\overline{B}\cap\overline{C}) = 2\Rightarrow n(B\cup C)=N - 2 \). So \( n(B\cap C)=(N - 6)+(N - 9)-(N - 2)=N - 13 \)

\( n(\overline{P}\cap\overline{C}) = 6\Rightarrow n(P\cup C)=N - 6 \). So \( n(P\cap C)=(N - 10)+(N - 9)-(N - 6)=N - 13 \)

\( n(\overline{P}\cap\overline{B}\cap\overline{C}) = 1\Rightarrow n(P\cup B\cup C)=N - 1 \)

And by inclusion - exclusion:

\( n(P\cup B\cup C)=n(P)+n(B)+n(C)-n(P\cap B)-n(P\cap C)-n(B\cap C)+n(P\cap B\cap C) \)

Substitute:

\( N - 1=(N - 10)+(N - 6)+(N - 9)-(N - 13)-(N - 13)-(N - 13)+1 \)

Simplify the right - hand side:

\( (N - 10)+(N - 6)+(N - 9)=3N-25 \)

\( -(N - 13)-(N - 13)-(N - 13)=-3N + 39 \)

So \( 3N-25-3N + 39+1=15 \)

Then \( N - 1 = 15\Rightarrow N = 16 \)

Wait, but let's check with the region method again.

Wait, maybe I made a mistake in the region definitions. Let's use a Venn diagram with three sets \( P \), \( B \), \( C \).

The number of people who don't go to any of the three is 1.

Let's define:

- Only \( P \): \( a \)
- Only \( B \): \( b \)
- Only \( C \): \( c \)
- \( P \) and \( B \) only: \( d \)
- \( P \) and \( C \) only: \( e \)
- \( B \) and \( C \) only: \( f \)
- All three: \( g = 1 \)
- None: \( h = 1 \)

We know:

1. \( n(\overline{P})=10 \): This is \( b + f+h + \) (only \( B \) and \( C \) and none? No, \( \overline{P} \) is all not in \( P \), so \( b + f+h + \) (only \( B \), only \( C \) and \( B \), none). Wait, \( \overline{P}= \) (only \( B \)) \( + \) (only \( C \) and \( B \)) \( + \) (only \( B \) and \( C \) and none? No, \( \overline{P}=\) (not in \( P \)) \(=\) (only \( B \)) \(+\) (only \( C \) and \( B \)) \(+\) (only \( C \)) \(+\) (none) \(+\) (only \( B \) and \( C \))? No, the correct breakdown of \( \overline{P} \) is:

\( \overline{P}=(\overline{P}\cap\overline{B}\cap\overline{C})+(\overline{P}\cap B\cap\overline{C})+(\overline{P}\cap\overline{B}\cap C)+(\overline{P}\cap B\cap C) \)

Which is \( h + b + c + f \)

We know \( h = 1 \), and \( n(\overline{P}) = 10\), so \( b + c + f+1 = 10\Rightarrow b + c + f=9 \)

2. \( n(\overline{B}) = 6\): \( \overline{B}=(\overline{P}\cap\overline{B}\cap\overline{C})+(\overline{P}\cap\overline{B}\cap C)+(P\cap\overline{B}\cap\overline{C})+(P\cap\overline{B}\cap C) \)

Which is \( h + c + a + e \)

\( 1 + c + a + e=6\Rightarrow a + c + e=5 \)

3. \( n(\overline{C}) = 9\): \( \overline{C}=(\overline{P}\cap\overline{B}\cap\overline{C})+(\overline{P}\cap B\cap\overline{C})+(P\cap\overline{B}\cap\overline{C})+(P\cap B\cap\overline{C}) \)

Which is \( h + b + a + d \)

\( 1 + b + a + d=9\Rightarrow a + b + d=8 \)

4. \( n(\overline{P}\cap\overline{B}) = 3\): \( \overline{P}\cap\overline{B}=(\overline{P}\cap\overline{B}\cap\overline{C})+(\overline{P}\cap\overline{B}\cap C) \)

Which is \( h + c \)

\( 1 + c=3\Rightarrow c = 2 \)

5. \( n(\overline{B}\cap\overline{C}) = 2\): \( \overline{B}\cap\overline{C}=(\overline{P}\cap\overline{B}\cap\overline{C})+(P\cap\overline{B}\cap\overline{C}) \)

Which is \( h + a \)

\( 1 + a=2\Rightarrow a = 1 \)

6. \( n(\overline{P}\cap\overline{C}) = 6\): \( \overline{P}\cap\overline{C}=(\overline{P}\cap\overline{B}\cap\overline{C})+(\overline{P}\cap B\cap\overline{C}) \)

Which is \( h + b \)

\( 1 + b=6\Rightarrow b = 5 \)

Now, from \( a + c + e=5 \), with \( a = 1 \) and \( c = 2 \), we get \( e=5-(1 + 2)=2 \)

From \( a + b + d=8 \), with \( a = 1 \) and \( b = 5 \), we get \( d=8-(1 + 5)=2 \)

From \( b + c + f=9 \), with \( b = 5 \) and \( c = 2 \), we get \( f=9-(5 + 2)=2 \)

Now, the total number of family members is \( a + b + c + d + e + f + g+h \)

\( a = 1 \), \( b = 5 \), \( c = 2 \), \( d = 2 \), \( e = 2 \), \( f = 2 \), \( g = 1 \), \( h = 1 \)

Sum: \( 1+5 + 2+2+2+2+1+1=16 \)

Answer:

\( 16 \)