Quantity I: n(s) = ¹⁶C₃ = (16× 15× 14)/(3× 2× 1) = 560 n(E) = ⁷C₃ = (7× 6× 5)/(3× 2× 1) = 35 P(E) = 35/560 = 35/560 = 1/16 Quantity II. n(s) = ¹⁶C₃ = (16× 15× 14)/(3× 2× 1) = 560 n(E) = ⁴C₁× ⁵C₁× ⁷C₁ P(E) = 140/560 = 1/4 Hence, Quantity I < Quantity II
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