If $s = {1, 2, 3, 4, 5, 6, 7, 8, 9}$ for $k = 1, 2, 3, 4, 5$ let $N_{k}$ be the number of sub sets of S each containing five elements out of which exactly K are odd, then $\sum_{k = 1}^{5} N_{k} =____$

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126

160

140

answer is C.

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Detailed Solution

$\begin{array}{l} N_{2} = n u m b e r o f s u b s e t o o f' S' e a c h c o n t a in g \\ 5 e l e m e n t s o u t o f w h i c h e x a c t l y K a r e o d d \\ i . e N_{1} = n o . o f 5 e l e m e n t s s u b s e t o c o n t a i n s' 1' o d d \\ = 5_{C_{1}} \times 4_{C_{4}} = 5 \\ N_{2} = n o . o f 5 e l e m e n t s s u b s e t o c o n t a i n s' 2' o d d n u m b e r \\ = 5_{C_{1}} \times 4_{C_{3}} = 40 \\ N_{3} = 5_{C_{3}} \times 4_{C_{2}} = 60 \\ N_{4} = 5_{C_{4}} \times 4_{C_{1}} = 20 \\ N_{5} = 5_{C_{5}} \times 4_{C_{0}} = 1 \\ \sum_{K = 1}^{5} N_{K} = N_{1} + N_{2} + N_{3} + N_{4} + N_{5} = 126 \end{array}$