The problem asks for the number of subsets of set $A = \{a, b, c, d\}$ that contain exactly 3 elements.

Solution:

To find the number of such subsets, we use the combination formula:

$C(n, r) = \frac{n!}{r!(n-r)!}$

Here:

• $n = 4$ (the number of elements in set $A$),
• $r = 3$ (the number of elements in each subset).

Substituting these values:

$C(4, 3) = \frac{4!}{3!(4-3)!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{(3 \cdot 2 \cdot 1)(1)} = 4$

Thus, the number of subsets of $A$ with exactly 3 elements is 4.

Explanation:

The subsets of $A$ containing 3 elements are:

1. $\{a, b, c\}$,
2. $\{a, b, d\}$,
3. $\{a, c, d\}$,
4. $\{b, c, d\}$.

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Questions for further exploration:

1. What is the total number of subsets of set $A$?
2. How many subsets of $A$ contain exactly 2 elements?
3. How many subsets of $A$ contain no elements (empty subset)?
4. How does the formula for combinations relate to Pascal's Triangle?
5. Can we generalize this process for larger sets with more elements?

Tip:

Remember, the sum of the binomial coefficients for a set of $n$ elements equals $2^n$, which represents the total number of subsets of the set!