Having problem with finding the number of ordered pairs.

Question

Y and Z are proper subset of X this means that X is having all the elements of Y and Z and also Y$\ne$X and Z$\ne$X (This is because we are talking about proper subset and not just subset).

Let Y={1} and Z={2,3,4,5} here the number of ordered pair is 4

Now let Y={2} and Z={1,3,4,5} here the number of ordered pair is 4

Now let Y={3} and Z={1,2,4,5} here the number of ordered pair is 4

Now let Y={4} and Z={1,2,3,5} here the number of ordered pair is 4

Now let Y={5} and Z={1,2,3,4} here the number of ordered pair is 4

I can only see these four possibility of Y and Z and according to me the total number of ordered pair is 4*5=20.

What I think is that if you take Y other than above say Y={2,3} then the ordered pair here which you get is already obtained in the above cases.

Answer in book is $3^5$ There is no explanation given in book.

Answer 1

The requirement that $Y \cap Z = \emptyset$ does not imply that $Y \cup Z = X$. For instance, $Y = \{1\}$ and $Z = \{2, 3\}$ are disjoint, but $Y \cup Z \neq X$. Also, note that the question does not require that $Y$ and $Z$ be proper subsets.

If the subsets $Y$ and $Z$ of $X = \{1, 2, 3, 4, 5\}$ are disjoint, each of the five element of $X$ is either in $Y$, in $Z$, or neither. Hence, there are $3^5$ ordered pairs of subsets $(Y, Z)$ such that $Y \subseteq X$, $Z \subseteq X$, and $X \cap Y = \emptyset$.

Answer 2

You can pick the set $Y$ of $k$ elements in $5 \choose k$ ways (any subset of $X$ with k elements).
For each such pick/choice of Y, you can then pick $Z$ in $2^{5-k}$ ways
(as a sub-set of the remaining $5-k$ elements from $X$).

So the total number of ways to form the couple $(Y,Z)$ is:

$\sum_{k=0}^5 {5 \choose k } \cdot 2^{5-k}$

This can be rewritten as:

$\sum_{k=0}^5 {5 \choose k } \cdot 1^k \cdot 2^{5-k}$

As we know from the binomial formula this is equal to $3^5$.