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6.Permutation and Combination
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If $^{2017}C_0 + ^{2017}C_1 + ^{2017}C_2+......+ ^{2017}C_{1008} = \lambda ^2 (\lambda > 0),$ then remainder when $\lambda $ is divided by $33$ is-
A
$8$
B
$13$
C
$17$
D
$25$
Solution
${\,^{2017}}{C_0} + {\,^{2017}}{C_1} + ……{\,^{2017}}{C_{1008}} = {2^{2016}} = {\lambda ^2}$
$\lambda = {2^{1008}}\,\,\,\,\, \Rightarrow \,\,\,{8.32^{201}} = 8{\left( {33 – 1} \right)^{201}} = – 8 = 25$
Standard 11
Mathematics
