Let S= 1,2,3, ldots ldots, 9 . For k=1,2, ldots ldots, 5, let Nk be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N1+N2+N3+N4+N5=

Question

Let S= 1,2,3, ldots ldots, 9 . For k=1,2, ldots ldots, 5, let Nk be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N1+N2+N3+N4+N5= (A) 210 (B) 252 (C) 125 (D) 126

Tardigrade Admin · Accepted Answer

There are only 4 even numbers in S ∴ Any subset of 5 elements of S will have at least 1 odd number. ⇒ N 1+ N 2+ N 3+ N 4+ N 5= 9 C 5=126

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

210

252

125

126

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

Q. Let S={1,2,3,……,9}. For k=1,2,……,5, let Nk​ be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N1​+N2​+N3​+N4​+N5​=

210

252

125

126

There are only 4 even numbers in S ∴ Any subset of 5 elements of S will have at least 1 odd number. ⇒N1​+N2​+N3​+N4​+N5​=9C5​=126

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

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