The problem asks to find the number of elements in the intersection of sets $P$ and $Q$, denoted as $n(P \cap Q)$. Set $P$ is defined as $\{x : 1 \le x \le 6\}$, where $x$ is an integer. Set $Q$ is defined as $\{x : 2 < x < 10\}$, where $x$ is an integer.

1. Problem Description

The problem asks to find the number of elements in the intersection of sets $P$ and $Q$, denoted as $n(P \cap Q)$. Set $P$ is defined as $\{x : 1 \le x \le 6\}$, where $x$ is an integer. Set $Q$ is defined as $\{x : 2 < x < 10\}$, where $x$ is an integer.

2. Solution Steps

First, we list the elements of set $P$. Since $x$ is an integer and $1 \le x \le 6$, we have $P = \{1, 2, 3, 4, 5, 6\}$.

Next, we list the elements of set $Q$. Since $x$ is an integer and $2 < x < 10$, we have $Q = \{3, 4, 5, 6, 7, 8, 9\}$.

Now, we find the intersection of sets $P$ and $Q$, which is $P \cap Q$. The intersection contains elements that are present in both sets. Comparing the elements of $P$ and $Q$, we have $P \cap Q = \{3, 4, 5, 6\}$.

Finally, we find the number of elements in the intersection $P \cap Q$. The set $P \cap Q$ contains 4 elements. Therefore, $n(P \cap Q) = 4$.

3. Final Answer

4