Sum of Subsets Containing At Least One Odd Element

To find the sum $N_1 + N_2 + N_3 + N_4 + N_5$, we analyze the set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

Step 1: Classification of elements in $S$ The set $S$ consists of 9 distinct elements:

• Odd elements: $\{1, 3, 5, 7, 9\}$ (Total of 5 elements)
• Even elements: $\{2, 4, 6, 8\}$ (Total of 4 elements)

Step 2: Understanding $N_k$ $N_k$ represents the number of 5-element subsets of $S$ containing exactly $k$ odd numbers. Since each subset contains 5 elements, if it has $k$ odd numbers, it must contain $(5-k)$ even numbers.

Using combinations, $N_k$ is given by: $N_k = \binom{5}{k} \times \binom{4}{5-k}$

Step 3: Calculating each $N_k$ for $k = 1, 2, 3, 4, 5$

• For $k = 1$: $N_1 = \binom{5}{1} \times \binom{4}{4} = 5 \times 1 = 5$

• For $k = 2$: $N_2 = \binom{5}{2} \times \binom{4}{3} = 10 \times 4 = 40$

• For $k = 3$: $N_3 = \binom{5}{3} \times \binom{4}{2} = 10 \times 6 = 60$

• For $k = 4$: $N_4 = \binom{5}{4} \times \binom{4}{1} = 5 \times 4 = 20$

• For $k = 5$: $N_5 = \binom{5}{5} \times \binom{4}{0} = 1 \times 1 = 1$

Step 4: Alternative Method (Total Subsets) Notice that the total number of 5-element subsets from $S$ is $\binom{9}{5}$. Since there are only 4 even elements in total, it is impossible to form a 5-element subset with 0 odd elements (i.e., $N_0 = \binom{5}{0} \times \binom{4}{5} = 0$).

Therefore, the sum $N_1 + N_2 + N_3 + N_4 + N_5$ is simply the total number of ways to choose 5 elements out of 9: $\sum_{k=1}^{5} N_k = \binom{9}{5} = \frac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1} = 126$

Summing the calculated individual values confirms this result: $N_1 + N_2 + N_3 + N_4 + N_5 = 5 + 40 + 60 + 20 + 1 = 126$

Thus, the correct option is D.