Let $A = \{a, b, c, d, e\}$. The no. of subsets that contain exactly 3 of these elements are?

The number of ways to select exactly 3 elements from the set $A = \{a, b, c, d, e\}$ is found using combinations since order doesn't matter.

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The combination formula:

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

where $n = 5$ (total elements) and $r = 3$ (elements to select).

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$C(5,3) = \dfrac{5!}{3!(5-3)!}$

$= \dfrac{5!}{3! \times 2!}$

$= \dfrac{5 \times 4 \times 3!}{3! \times 2 \times 1}$

$= \dfrac{5 \times 4}{2 \times 1}$

$= \dfrac{20}{2}$

$= 10$

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The 10 subsets are: $\{a,b,c\}$, $\{a,b,d\}$, $\{a,b,e\}$, $\{a,c,d\}$, $\{a,c,e\}$, $\{a,d,e\}$, $\{b,c,d\}$, $\{b,c,e\}$, $\{b,d,e\}$, $\{c,d,e\}$

Therefore, there are $10$ subsets that contain exactly $3$ elements.