Explanation
Exponent of $$18$$ in $$200!$$
$$\Rightarrow 18=(3)^{ 2 }\times 2$$
Exponent of $$P$$ in $$n!$$
$$=\left[ \dfrac { n }{ p } \right] +\left[ \dfrac { n }{ p^{ 2 } } \right] +.............\left[ \dfrac { n }{ p^{ K } } \right] \\ =P^{ K }\le n$$
Exponential of $$3$$ in $$n!$$
$$=\left[ \dfrac { 200 }{ 3 } \right] +\left[ \dfrac { 200 }{ 9 } \right] +\left[ \dfrac { 200 }{ 27 } \right] +\left[ \dfrac { 200 }{ 81 } \right] \\ =66+22+4+2\\ =94$$
Exponent of $$(3)^{ 2 }=48$$
Exponent of $$2$$ in $$n!$$
$$=\left[ \dfrac { 200 }{ 2 } \right] +\left[ \dfrac { 200 }{ 4 } \right] +\left[ \dfrac { 200 }{ 8 } \right] +\left[ \dfrac { 200 }{ 16 } \right] >>48$$
Least exponent $$=48$$
Exponential of $$18$$ is $$48$$
Hence the correct answer is $$48$$
Out of $$8 $$ men and $$10$$ women . The committee has to select $$5$$ men and $$6$$ women.
$$^8C_5 \times ^{10}C_6 =11760$$
$$2$$ White balls
$$3$$ Black balls
$$4$$ Red balls
Three balls are drawn
Total possible combination
$$\Rightarrow ^{ 9 }{ P }_{ 3 }$$
Combination without black
$$\Rightarrow ^{ 2 }{ C_{ 2 } }\times ^{ 4 }{ C }_{ 1 }+^{ 2 }{ C }_{ 1 }\times ^{ 4 }{ C }_{ 2 }+^{ 4 }{ C }_{ 3 }$$
The combination with atleast one black
$$\Rightarrow \dfrac { 9\times 8\times 7 }{ 3\times 2\times 1 } -\left[ 1\times 4+2\times \dfrac { 12 }{ 2 } +4 \right] \\ \Rightarrow 12(7)-\left[ 20 \right] \\ \Rightarrow 84-20=64$$
$$64$$ Combination include atleast one black
Hence the correct answer is $$64$$
Please disable the adBlock and continue. Thank you.