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 =

Aicharya · Accepted Answer

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…

Solution

Let S = 1,2 , 3 , &#8230; . 9 . For k = 1,2 , &#8230; 5 , let N k be the number of subsets of S , each…

Solution

Let S = 1,2 , 3 , … . 9 . For k = 1,2 , … 5 , let N k be the number of subsets of S , each…