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Question:

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 = ?

125

126

210

252

Solution:

In the given set1,3,5,7,9are odd and2,4,6,8are evenTherefore there are5odd and4even numbers in the setN1=5C1×4C4=51 odd and 4 evenN2=5C2×4C3=402 odd and 3 evenN3=5C3×4C2=603 odd and 2 evenN4=5C1×4C4=204 odd and 1 evenN5=5C5×4C0=15 odd and 0 evenThereforeN1+N2+N3+N4+N5=5+40+60+20+1=126