Correct Option: C Let O denote the set of five distinct odd integers, and E denote the set of five distinct even integers. The total number of integers is 10. **Step 1: Calculate the total number of possible outcomes (selection and arrangement).** We choose 4 integers from 10, and then arrange them. This is equivalent to finding the number of permutations of 10 items taken 4 at a time. Total possible outcomes = $P(10,4) = \frac{10!}{(10-4)!} = 10 \times 9 \times 8 \times 7 = 5040$. **Step 2: Calculate the number of favorable outcomes (selection and arrangement such that no two odd integers are adjacent).** Let $n_O$ be the number of odd integers chosen and $n_E$ be the number of even integers chosen, such that $n_O + n_E = 4$. We consider different compositions of the chosen set: * **Case 1: $n_O = 0, n_E = 4$ (0 Odd, 4 Even)** * Number of ways to choose 0 odd from 5 and 4 even from 5: $\binom{5}{0} \times \binom{5}{4} = 1 \times 5 = 5$ ways. * Arrangement: Since there are no odd integers, the condition 'no two odd integers are adjacent' is trivially satisfied. The 4 distinct even integers can be arranged in $4! = 24$ ways. * Favorable outcomes for Case 1: $5 \times 24 = 120$. * **Case 2: $n_O = 1, n_E = 3$ (1 Odd, 3 Even)** * Number of ways to choose 1 odd from 5 and 3 even from 5: $\binom{5}{1} \times \binom{5}{3} = 5 \times 10 = 50$ ways. * Arrangement: With only one odd integer, it cannot be adjacent to another odd integer. All arrangements are valid. The 4 distinct integers can be arranged in $4! = 24$ ways. * Favorable outcomes for Case 2: $50 \times 24 = 1200$. * **Case 3: $n_O = 2, n_E = 2$ (2 Odd, 2 Even)** * Number of ways to choose 2 odd from 5 and 2 even from 5: $\binom{5}{2} \times \binom{5}{2} = 10 \times 10 = 100$ ways. * Arrangement: To ensure no two odd integers are adjacent, we use the 'gap method': 1. Arrange the 2 distinct even integers first: $2! = 2$ ways. 2. These arrangements create 3 possible positions (gaps) for the odd integers (e.g., $\_ E \_ E \_$). 3. Choose 2 of these 3 positions for the 2 distinct odd integers: $\binom{3}{2} = 3$ ways. 4. Arrange the 2 distinct odd integers in the chosen positions: $2! = 2$ ways. So, the number of favorable arrangements for two odd and two even integers = $2! \times \binom{3}{2} \times 2! = 2 \times 3 \times 2 = 12$ ways. * Favorable outcomes for Case 3: $100 \times 12 = 1200$. * **Case 4: $n_O = 3, n_E = 1$ (3 Odd, 1 Even)** * Number of ways to choose 3 odd from 5 and 1 even from 5: $\binom{5}{3} \times \binom{5}{1} = 10 \times 5 = 50$ ways. * Arrangement: With 3 odd integers and 1 even integer, it is impossible to arrange them such that no two odd integers are adjacent. If the even integer is placed, it creates two gaps (e.g., $\_ E \_$). Placing 3 odd integers in these 2 gaps will always result in at least two odd integers being adjacent. * Favorable arrangements for Case 4 = 0. * Favorable outcomes for Case 4: $50 \times 0 = 0$. * **Case 5: $n_O = 4, n_E = 0$ (4 Odd, 0 Even)** * Number of ways to choose 4 odd from 5 and 0 even from 5: $\binom{5}{4} \times \binom{5}{0} = 5 \times 1 = 5$ ways. * Arrangement: With 4 odd integers, it is impossible to arrange them such that no two are adjacent. * Favorable arrangements for Case 5 = 0. * Favorable outcomes for Case 5: $5 \times 0 = 0$. **Step 3: Sum the favorable outcomes and calculate the probability.** Total favorable outcomes = $120 + 1200 + 1200 + 0 + 0 = 2520$. Probability = $\frac{\text{Total Favorable Outcomes}}{\text{Total Possible Outcomes}} = \frac{2520}{5040} = \frac{1}{2}$. Thus, the probability that no two odd integers are adjacent in the arrangement is $\frac{1}{2}$.