JEE Advance - Mathematics (2017 - Paper 2 Offline - No. 5)
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
126
125
Explanation
$${N_i} = {}^5{C_k} \times {}^4{C_{5 - k}}$$
$$\eqalign{ & {N_1} = 5 \times 1 \cr & {N_2} = 10 \times 4 \cr & {N_3} = 10 \times 6 \cr & {N_4} = 5 \times 4 \cr & {N_5} = 1 \cr & {N_1} + {N_2} + {N_3} + {N_4} + {N_5} = 126 \cr} $$
$$\eqalign{ & {N_1} = 5 \times 1 \cr & {N_2} = 10 \times 4 \cr & {N_3} = 10 \times 6 \cr & {N_4} = 5 \times 4 \cr & {N_5} = 1 \cr & {N_1} + {N_2} + {N_3} + {N_4} + {N_5} = 126 \cr} $$
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