Let S = 1,2 , 3 , … . 9 . For k = 1,2 , … 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 =

Question

Let S = 1,2 , 3 , &#8230; . 9 . For k = 1,2 , &#8230; 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 =

Quantrex Academy · Accepted Answer

Correct answer: D. 126

N 1 + N 2 + N 3 + N 4 + N 5 = Total ways &ndash; when no odd Total ways = 9 C 5 Number of ways when no odd, is zero ( &#8757; only available even are 2 , &#160; 4 , &#160; 6 , &#160; 8 ) &#8756; &#160; 9 C 5 - zero = 126

Let S = 1,2 , 3 , … . 9 . For k = 1,2 , … 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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More from Sets and Relations

Let S = 1,2 , 3 , &#8230; . 9 . For k = 1,2 , &#8230; 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 =

Options

Correct answer

Step-by-step solution

More from Sets and Relations

Let S = 1,2 , 3 , … . 9 . For k = 1,2 , … 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 =