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JEE Advanced 2017Paper 2

Exam: JEE Advanced
Year: 2017
Paper: Paper 2
Subject: Mathematics
Chapter: Sets and Relations

JEE Advanced 2017 · Paper 2 · Mathematics · Sets and Relations

Let S = \ 1,2, 3, . 9\ . For k = 1,2, 5, let N _ k be the number of...

Let

S = \left\{1,2, 3, \ldots . 9\right\}

. For

k = 1,2, \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} =

Options

A $125$
B $252$
C $210$
D $126$

Correct answer

D. $126$

Step-by-step solution

${N}_{1} + {N}_{2} + {N}_{3} + {N}_{4} + {N}_{5} =$ Total ways – {when no odd}

Total ways ${=}^{9}{C}_{5}$

Number of ways when no odd, is zero ( $∵$ only available even are $2, 4, 6, 8$ )

$∴ {}^{9}{C}_{5} - zero = 126$

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