Let S = {1, 2, 3, ldots, 9}. For k = 1, 2, ld | vedclass

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(D) The set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ contains $5$ odd numbers $\{1, 3, 5, 7, 9\}$ and $4$ even numbers $\{2, 4, 6, 8\}$.We are looking for

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$126$

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(D) The set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ contains $5$ odd numbers $\{1, 3, 5, 7, 9\}$ and $4$ even numbers $\{2, 4, 6, 8\}$.We are looking for the total number of subsets of $S$ with $5$ elements,where the number of odd elements $k$ can be $1, 2, 3, 4,$ or $5$.Since there are only $4$ even numbers in $S$,any subset of $5$ elements must contain at least $5 - 4 = 1$ odd number.Therefore,the sum $N_1 + N_2 + N_3 + N_4 + N_5$ represents the total number of ways to choose $5$ elements from the $9$ elements in $S$.This is given by the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.$N_1 + N_2 + N_3 + N_4 + N_5 = \binom{9}{5} = \binom{9}{4} = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$.

Let $S = \{1, 2, 3, \ldots, 9\}$. For $k = 1, 2, \ldots, 5$,let $N_k$ be the number of subsets of $S$,each containing five elements out of which exactly $k$ are odd. Then $N_1 + N_2 + N_3 + N_4 + N_5 =$

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