Ask Questions, Get Answers

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

Given 5 different green dyes ,four different blue dyes and three different red dyes,the number of combinations of dyes which can be chosen taking at least one green and one blue dye is

$\begin{array}{1 1}(A)\;3600\\(B)\;3720\\(C)\;3800\\(D)\;3600\end{array} $

Can you answer this question?

1 Answer

0 votes
Green dyes =5
Blue dyes =4
Red dyes =3
The no of ways of choosing of at least one green and one blue is =$(2^5-1)\times (2^4-1)\times 2^3$
$\Rightarrow (32-1)\times (16-1)\times 8$
$\Rightarrow 31\times 15\times 8$
$\Rightarrow 3720$
Hence (B) 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