10 different vectors are lying on a plane out of which four are parallel with respect…

MATHEMATICS · Single correct · CBSE · Mathematics · Permutations and Combinations · Permutations and Combinations

Problem

10 different vectors are lying on a plane out of which four are parallel with respect to each other. Probability that three vectors chosen from them will satisfy the equation λ1a+λ2b+λ3c=0\lambda_1\mathbf{a} + \lambda_2\mathbf{b} + \lambda_3\mathbf{c} = 0 , where λ1,λ2\lambda_1, \lambda_2 and λ3≠0\lambda_3 \neq 0 is (1) 6C2×4C110C3\frac{{}^6C_2 \times {}^4C_1}{{}^{10}C_3} (2) (6C3×4C1)+6C310C3\frac{({}^6C_3 \times {}^4C_1) + {}^6C_3}{{}^{10}C_3} (3) (6C2×4C1)+(4C3)10C3\frac{({}^6C_2 \times {}^4C_1) + ({}^4C_3)}{{}^{10}C_3} (4) (6C2+4C1)+(6C2×4C1)10C3\frac{({}^6C_2 + {}^4C_1) + ({}^6C_2 \times {}^4C_1)}{{}^{10}C_3}

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