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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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A five digit number divisible by 3 is to be formed using the numbers 0,1,2,3,4 and 5 without repetitions.The total number of ways this can be done is

$\begin{array}{1 1}(A)\;216\\(B)\;600\\(C)\;240\\(D)\;3125\end{array} $

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1 Answer

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  • $n!=n(n-1)(n-2)(n-3).......(3)(2)(1)$
1+2+3+4+5=15 divisible by 3
The no of five digit numbers formed =5! ways
$\Rightarrow 120$ ways
The no of five digit numbers that can be formed using 0,1,2,3,4 can be
First digit $\rightarrow 4$ways
Second digit $\rightarrow 4$ways
Third digit $\rightarrow 3$ways
Fourth digit $\rightarrow 2$ways
Fifth digit $\rightarrow 1$way
Total no of ways =$4\times 4\times 3\times 2\times 1$
$\Rightarrow 4\times 4!$
$\Rightarrow 96$
Hence total number of ways =$120+96$
$\Rightarrow 216$
Hence (A) is the correct answer.
answered May 16, 2014 by sreemathi.v

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