JEE Advanced 2017 · Paper 2 · Mathematics Q6

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MathematicsSingle correct+3 / -1Question 42 of 54

Let $S = \{1, 2, 3, \ldots, 9\}$ . 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 $N_1 + N_2 + N_3 + N_4 + N_5 =$

210

252

125

126

Answer and worked solution

Official answer: D

$S$ has 5 odd and 4 even numbers. A 5-element subset cannot be made of even numbers alone, so it always has at least one odd element.

So $N_1+\dots+N_5$ counts every 5-element subset:
$\binom95=126.$

(D)

Topics:Combinations

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JEE Advanced 2017 · Paper 2, Mathematics question 6