logo

Ask Questions, Get Answers

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

The number of ways of distributing 8 identical balls in 3 distinct boxes,so that none of the boxes is empty is

$\begin{array}{1 1}(A)\;5\\(B)\;21\\(C)\;3^8\\(D)\;8C_3\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
  • $n!=n(n-1)(n-2).....(3)(2)(1)$
The total no of ways of dividing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is
$(8-1)C_{3-1}=7C_2$
$\Rightarrow \large\frac{7!}{2!5!}$
$\Rightarrow \large\frac{7\times 6\times 5!}{2\times 5!}$
$\Rightarrow 21$
Hence (B) is the correct answer.
answered Jun 20, 2014 by sreemathi.v
 

Related questions

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