Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
0 votes

The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women and men is

$\begin{array}{1 1}(A)\;94\\(B)\;126\\(C)\;128\\(D)\;\text{None}\end{array} $

Can you answer this question?

1 Answer

0 votes
  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Total no of men =4
Total no of women =6
The total no of ways =$4C_2\times 6C_4+4C_3\times 6C_6$
$4C_2=\large\frac{4!}{2!2!}=\frac{4\times 3\times 2!}{2\times 2!}$$=6$
$6C_4=\large\frac{6!}{4!2!}=\frac{6\times 5\times 4!}{4!\times 2}$$=15$
$4C_3=\large\frac{4!}{3!1!}=\frac{4\times 3!}{3!\times 1}$$=4$
$4C_2\times 6C_4+4C_3\times 6C_6=6\times 15+4\times 1$
$\Rightarrow 90+4$
$\Rightarrow 94$
Hence (A) is the correct answer.
answered May 16, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App