In how many ways can a committee of 5 members be selected from 6 men and 5 ladies- consisting of 3 men and 2 ladies-

Question

Toppr Admin · Accepted Answer

Correct option is D- -200-Number of ways of selecting -3- men out of -6- and -2- ladies out of -5-6C-3-times -5C-2-left-dfrac-6-times 5-times 4-3-times 2-times 1-times -dfrac-5-times 4-2-times 1-right-200-

In how many ways can a committee of $$5$$ members be selected from $$6$$ men and $$5$$ ladies, consisting of $$3$$ men and $$2$$ ladies?

Correct option is D. $$200$$
Number of ways of selecting $$3$$ men out of $$6$$ and $$2$$ ladies out of $$5$$
$$=(^6C_3\times {^5C_2})=\left(\dfrac{6\times 5\times 4}{3\times 2\times 1}\times \dfrac{5\times 4}{2\times 1}\right)=200$$.

In how many ways can a committee of $$5$$ members be selected from $$6$$ men and $$5$$ ladies, consisting of $$3$$ men and $$2$$ ladies?

Correct option is D. $$200$$Number of ways of selecting $$3$$ men out of $$6$$ and $$2$$ ladies out of $$5$$$$=(^6C_3\times {^5C_2})=\left(\dfrac{6\times 5\times 4}{3\times 2\times 1}\times \dfrac{5\times 4}{2\times 1}\right)=200$$.

Correct option is D. $$200$$
Number of ways of selecting $$3$$ men out of $$6$$ and $$2$$ ladies out of $$5$$
$$=(^6C_3\times {^5C_2})=\left(\dfrac{6\times 5\times 4}{3\times 2\times 1}\times \dfrac{5\times 4}{2\times 1}\right)=200$$.