Find the intersection value of 3 subsets

[schinb65]

Let a, b and c be three subsets of universe U with the following properties: n(A)= 63, n(B)=91, n(c)=44, The intersection of (A&B)= 25, The intersection of (A&C)=23, The intersection of (C&B)=21, n(A U B U C)= 139. Find the intersection of (A&B&C).

I am told the answer is 10. I tried drawing a diagram.

A: B: C:
x x x
25-x 21-x 23-x
23-x 25-x 21-x
63-(x + 21-x + 23-x) 91-(x+25-x+21-x) 44-(x+23-x+21-x)I added all of these together and had them equal to 198, since I am using values repeatedly. I also omitted the repeats and set the sum to 139. This should give me the correct answer correct?

 

[caffeinemachine]

Let a, b and c be three subsets of universe U with the following properties: n(A)= 63, n(B)=91, n(c)=44, The intersection of (A&B)= 25, The intersection of (A&C)=23, The intersection of (C&B)=21, n(A U B U C)= 139. Find the intersection of (A&B&C).

I am told the answer is 10. I tried drawing a diagram.

A: B: C:
x x x
25-x 21-x 23-x
23-x 25-x 21-x
63-(x + 21-x + 23-x) 91-(x+25-x+21-x) 44-(x+23-x+21-x)I added all of these together and had them equal to 198, since I am using values repeatedly. I also omitted the repeats and set the sum to 139. This should give me the correct answer correct?

What you have written in the last paragraph is essentially what you need to do. I don't understand the rest of your attempt though but I think you are on the right track. You might find this useful Inclusion

 

[soroban]

Hello, schinb65!

Let $A, B, C$ be three subsets of universe $U$ with the following properties:

. . $\begin{array}{c}n(A)\,=\, 63 \\ n(B)\,=\,91 \\ n(C)\,=\,44\end{array} \qquad \begin{array}{c} n(A\cap B) \,=\,25 \\ n(B\cap C) \,=\,21 \\ n(A\cap C) \,=\, 23 \end{array} \qquad n(A\cup B \cup C) \,=\,139 $

Find: $ n(A \cap B\cap C)$

Are you familiar with this formula?

$n(A \cup B \cup C) \:=\:n(A) + n(B) + n(C) $

. . . . . . . . . . . . . . $+\,n(A \cap B) + n(B \cap C) + n(A \cap C) $

. . . . . . . . . . . . . . . . $+ n(A \cap B \cap C)$

 

[Jameson]

Hello, schinb65!


Are you familiar with this formula?

$n(A \cup B \cup C) \:=\:n(A) + n(B) + n(C) $

. . . . . . . . . . . . . . $+\,n(A \cap B) + n(B \cap C) + n(A \cap C) $

. . . . . . . . . . . . . . . . $+ n(A \cap B \cap C)$

Hi soroban,

I think what you meant was this:
$n(A \cup B \cup C) \:=\:n(A) + n(B) + n(C) -\,n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$.

Just a typo I think :)

Jameson

 

[blue_raver22]

Yes, your approach is correct. By setting the sum of all the elements in the subsets equal to the total number of elements in the universe, you can solve for the value of x, which represents the intersection of (A&B&C). In this case, x=10, as you correctly stated. This means that there are 10 elements that are common to all three subsets A, B, and C.