Question

Let $S=\\{1,2,3, \ldots \ldots, 9\\}$. For $k=1,2, \ldots \ldots, 5$, let $N_{k}$ be the number of subsets of $S$, each containing five elements out of which exactly $k$ are odd. Then $N_{1}+N_{2}+N_{3}+N_{4}+N_{5}=$

• A. 210
• B. 252
• C. 125
• D. 126

Answer 1

Answer: (D)

126

Answer 2

There are only $4$ even numbers in $S$ $\therefore$ Any subset of $5$ elements of $S$ will have at least $1$ odd number. $\Rightarrow N _{1}+ N _{2}+ N _{3}+ N _{4}+ N _{5}={ }^{9} C _{5}=126$