Let the set $$P = \{2,3,4, \dots, 25\}$$. For each $$k \in P$$, define $$Q(k) = \{x \in P$$ such that $$x > k$$ and $$k$$ divides $$x\}$$. Then the number of elements in the set $$P - {\bigcup}_{k = 2}^{25} Q(k)$$ is

$$P=\left\{2,3,4,...,25\right\}$$

Q(k)= {x ∈ P such that x > k and k divides x}. Now, we will have multiple values of k.

Case 1: When k = 2, Q(2) = {4,6,8,10,....,24}

Case 2: When k = 3, Q(3) = {6,9,12,15,....,24}

Case 3: When k = 4, Q(4) = {8,12,16,....24}

Case 4: When k = 5, Q(5) = {10,15,20,25}

Now, in the above 4 cases, we can see that - 2, 3, and 5 are not included, and in the next values of k, since each value of Q(k) is greater than k, thus 2, 3 and 5 will never be included. Now, in further values of k, the numbers which are prime will never be included in any set, because in Q(k), each element is such that x > k, and k divides x. But prime number is divisible by only 1 and itself. And the prime number cannot include itself as well, because x > k. Thus, $$U_{k=2}^{25}Q\left(k\right)$$ will contains all the numbers from 2 to 24, which are not prime. Thus, -

$$U_{k=2}^{25}Q\left(k\right)=Q\left(2\right)∪Q\left(3\right)∪Q\left(4\right)∪.....∪Q\left(25\right)$$

$$U_{k=2}^{25}Q\left(k\right)=\left\{4,6,8,9,10,12,14,15,16,18,20,21,22,24,25\right\}$$

$$P-U_{k=2}^{25}Q\left(k\right)=\left\{2,3,5,7,11,13,17,19,23\right\}$$

Thus, 9 elements are there in the above set.