1. Understand the Set S

The given set is S = {1, 2, 3, 4, 5, 6, 7, 8, 9}. We need to analyze the composition of this set in terms of odd and even numbers.

• Odd numbers in S: {1, 3, 5, 7, 9}. There are 5 odd numbers.
• Even numbers in S: {2, 4, 6, 8}. There are 4 even numbers.

2. Define Nk

Nk is defined as the number of subsets of S containing five elements, out of which exactly k are odd.

A 5-element subset is formed by selecting some odd and some even numbers from S. If a subset has k odd numbers, it must have 5 - k even numbers to make a total of 5 elements.

To find Nk, we need to:

1. Choose k odd numbers from the 5 available odd numbers. This can be done in C(5, k) ways.
2. Choose 5 - k even numbers from the 4 available even numbers. This can be done in C(4, 5 - k) ways.

So, the formula for Nk is: $N_k = \binom{5}{k} \times \binom{4}{5-k}$

3. Analyze the required sum

We need to find the value of N1 + N2 + N3 + N4 + N5. This sum represents the total number of 5-element subsets of S that have at least one odd number.

Let's consider all possible cases for the number of odd elements (k) in a 5-element subset:

• The number of odd elements, k, can be chosen from the 5 available odd numbers, so $0 \le k \le 5$.
• The number of even elements, 5-k, must be chosen from the 4 available even numbers, so $0 \le 5-k \le 4$.

From the second condition, $5-k \le 4$, which implies $k \ge 1$. Also, $5-k \ge 0$, which implies $k \le 5$. Combining these, the possible values for k are 1, 2, 3, 4, 5.

This means that it is impossible to form a 5-element subset of S with k=0 odd numbers, as it would require choosing 5 even numbers, but there are only 4 available.

Therefore, any 5-element subset of S must contain at least one odd number. The sum N1 + N2 + N3 + N4 + N5 accounts for all possible 5-element subsets of S.

4. Calculate the total number of 5-element subsets

The required sum is simply the total number of ways to choose 5 elements from the 9 elements in set S. This is given by the combination formula C(9, 5).

$C(9, 5) = \binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!}$

$C(9, 5) = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5! \times 4 \times 3 \times 2 \times 1} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}$

$C(9, 5) = 9 \times 2 \times 7 = 126$

Thus, N1 + N2 + N3 + N4 + N5 = 126.

5. Verification (Optional but Recommended)

We can also calculate each Nk individually and sum them up.

• N1 (k=1): 1 odd, 4 even N1 = C(5, 1) * C(4, 4) = 5 * 1 = 5

• N2 (k=2): 2 odd, 3 even N2 = C(5, 2) * C(4, 3) = 10 * 4 = 40

• N3 (k=3): 3 odd, 2 even N3 = C(5, 3) * C(4, 2) = 10 * 6 = 60

• N4 (k=4): 4 odd, 1 even N4 = C(5, 4) * C(4, 1) = 5 * 4 = 20

• N5 (k=5): 5 odd, 0 even N5 = C(5, 5) * C(4, 0) = 1 * 1 = 1

Sum: N1 + N2 + N3 + N4 + N5 = 5 + 40 + 60 + 20 + 1 = 126

Both methods yield the same result. The result matches option C.

(This sum is also an application of Vandermonde's Identity: \sum_{k=0}^{r} \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$. Here, `m=5`, `n=4`, `r=5`. The sum is \sum_{k=1}^{5} \binom{5}{k}\binom{4}{5-k}$. Since$\binom{4}{5-0} = \binom{4}{5} = 0, the sum is equal to $$\sum_{k=0}^{5} \binom{5}{k}\binom{4}{5-k} = \binom{5+4}{5} = \binom{9}{5} = 126.)