Which is the simplified representation of `(A' cap B' cap C) cup (B cap C ) cup (A cap C)` where A,B and C are subsets of set X (a).`A` (b). `B` (c).`C` (d).`X cap (A cup B cup C)`

Options

A. A
B. B
C. C
D. `X cap (A cup B cup C)`

Explanation

To simplify the expression \( (A' \cap B' \cap C) \cup (B \cap C) \cup (A \cap C) \), where \( A, B, \) and \( C \) are subsets of a universal set \( X \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the original expression:
\[ (A' \cap B' \cap C) \cup (B \cap C) \cup (A \cap C) \]

### Step 2: Apply De Morgan's Law
Using De Morgan's Law, we can rewrite \( A' \cap B' \) as \( (A \cup B)' \). Thus, we can rewrite the first part of the expression:
\[ (A' \cap B' \cap C) = ((A \cup B)' \cap C) \]

So the expression becomes:
\[ ((A \cup B)' \cap C) \cup (B \cap C) \cup (A \cap C) \]

### Step 3: Factor out \( C \)
Notice that \( C \) is common in all terms. We can factor \( C \) out:
\[ C \cap ((A \cup B)' \cup B \cup A) \]

### Step 4: Simplify the inner expression
Now we simplify the expression inside the parentheses:
\[ (A \cup B)' \cup B \cup A \]
Using the property that \( (A \cup B)' \cup A = X \) (the universal set), we can simplify:
\[ (A \cup B)' \cup A = X \]
Thus:
\[ X \cup B = X \]

### Step 5: Final expression
Now substituting back, we have:
\[ C \cap X = C \]

### Conclusion
The simplified representation of the original expression is:
\[ C \]

Thus, the answer is:
**(c) C**