the venn diagram here shows the cardinality of each set. use this to find the cardinality of the given set.
$n(a\cup b)=\square$

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# Explanation:
## Step1: Recall the formula for $ n(A \cup B) $
The formula for the cardinality of the union of two sets $ A $ and $ B $ is $ n(A \cup B)=n(A)+n(B)-n(A \cap B) $. First, we need to find $ n(A) $, $ n(B) $, and $ n(A \cap B) $ from the Venn diagram.

## Step2: Calculate $ n(A) $
To find $ n(A) $, we sum the numbers in the regions of set $ A $. The regions of $ A $ are: the part only in $ A $ (16), the part in $ A \cap B $ (6), and the part in $ A \cap C $ (4), and the part in $ A \cap B \cap C $ (1). Wait, no, actually, for set $ A $, the regions are: only $ A $ (16), $ A \cap B $ (6 + 1), $ A \cap C $ (4 + 1)? Wait, no, let's look at the Venn diagram. The Venn diagram has three sets $ A $, $ B $, $ C $. The regions:

- Only $ A $: 16
- $ A \cap B $ only (not including $ C $): 6
- $ A \cap C $ only (not including $ B $): 4
- $ A \cap B \cap C $: 1
- Only $ B $: 11
- $ B \cap C $ only (not including $ A $): 2
- Only $ C $: 15

Wait, maybe a better way: $ n(A) $ is the sum of all regions inside $ A $. So inside $ A $: only $ A $ (16), $ A \cap B $ (6 + 1), $ A \cap C $ (4 + 1). Wait, no, the intersection of $ A $ and $ B $ is the region where $ A $ and $ B $ overlap, which includes the part with $ C $ too? Wait, no, the standard Venn diagram for three sets:

- Region 1: Only $ A $: 16
- Region 2: $ A \cap B $ only (no $ C $): 6
- Region 3: Only $ B $: 11
- Region 4: $ B \cap C $ only (no $ A $): 2
- Region 5: Only $ C $: 15
- Region 6: $ A \cap C $ only (no $ B $): 4
- Region 7: $ A \cap B \cap C $: 1

So, $ n(A) $ is the sum of regions 1, 2, 6, 7. So $ n(A)=16 + 6+4 + 1=27 $.

## Step3: Calculate $ n(B) $
$ n(B) $ is the sum of regions inside $ B $: region 2, 3, 4, 7. So $ n(B)=6 + 11+2 + 1=20 $.

## Step4: Calculate $ n(A \cap B) $
$ n(A \cap B) $ is the sum of regions where $ A $ and $ B $ overlap, which are regions 2 and 7. So $ n(A \cap B)=6 + 1=7 $.

## Step5: Apply the union formula
Now, using $ n(A \cup B)=n(A)+n(B)-n(A \cap B) $. We have $ n(A)=27 $, $ n(B)=20 $, $ n(A \cap B)=7 $. Wait, wait, let's recalculate $ n(A) $ and $ n(B) $ again. Wait, maybe I made a mistake. Let's re-express:

Wait, the regions:

- Only $ A $: 16
- $ A \cap B $ (excluding $ C $): 6
- $ A \cap C $ (excluding $ B $): 4
- $ A \cap B \cap C $: 1
- Only $ B $: 11
- $ B \cap C $ (excluding $ A $): 2
- Only $ C $: 15

So $ n(A) $ is only $ A $ + $ A \cap B $ (including $ C $) + $ A \cap C $ (including $ C $)? No, no. The correct way: $ n(A) $ is the number of elements in $ A $, which is the sum of all elements in the circle labeled $ A $. So the circle $ A $ includes:

- Only $ A $: 16
- $ A \cap B $ (the part where $ A $ and $ B $ overlap, regardless of $ C $): 6 (only $ A \cap B $) + 1 ( $ A \cap B \cap C $)
- $ A \cap C $ (the part where $ A $ and $ C $ overlap, regardless of $ B $): 4 (only $ A \cap C $) + 1 ( $ A \cap B \cap C $)

Wait, no, that's double - counting the $ A \cap B \cap C $ region. The correct way is that the elements in $ A $ are:

Only $ A $: 16

$ A \cap B $ only (not in $ C $): 6

$ A \cap C $ only (not in $ B $): 4

$ A \cap B \cap C $: 1

So $ n(A)=16 + 6+4 + 1=27 $

Similarly, $ n(B) $ is:

Only $ B $: 11

$ A \cap B $ only (not in $ C $): 6

$ B \cap C $ only (not in $ A $): 2

$ A \cap B \cap C $: 1

So $ n(B)=11 + 6+2 + 1=20 $

$ n(A \cap B) $ is the number of elements in the intersection of $ A $ and $ B $, which is $ A \cap B $ only (6) plus $ A \cap B \cap C $ (1), so $ n(A \cap B)=6 + 1=7 $

Now, $ n(A \cup B)=n(A)+n(B)-n(A \cap B)=27 + 20-7=40 $

Wait, but let's check another way. Alternatively, we can sum all the regions in $ A $ or $ B $. The regions in $ A $ or $ B $ are:

Only $ A $: 16

$ A \cap B $ only (6)

$ A \cap B \cap C $ (1)

Only $ B $: 11

$ A \cap C $ only (4)

$ B \cap C $ only (2)? Wait, no, $ A \cup B $ includes all elements in $ A $ or $ B $, so the regions are:

- Only $ A $: 16
- $ A \cap B $ only (6)
- $ A \cap B \cap C $ (1)
- Only $ B $: 11
- $ A \cap C $ only (4) (since it's in $ A $)
- $ B \cap C $ only (2) (since it's in $ B $)? Wait, no, $ B \cap C $ only is in $ B $, so yes. Wait, no, $ A \cup B $ is all elements in $ A $ or $ B $, so:

Only $ A $: 16

$ A \cap B $ only (6)

$ A \cap B \cap C $ (1)

Only $ B $: 11

$ A \cap C $ only (4) (because it's in $ A $)

$ B \cap C $ only (2) (because it's in $ B $)

Now sum these: $ 16+6 + 1+11+4 + 2=40 $. Yep, that matches. So the cardinality of $ A \cup B $ is 40.

# Answer:
$ \boxed{40} $

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the formula for \( n(A \cup B) \)

Step2: Calculate \( n(A) \)

Step3: Calculate \( n(B) \)

Step4: Calculate \( n(A \cap B) \)

Step5: Apply the union formula

Answer: