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CBSE XI
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Permutations and Combinations
A five digit number divisible by 3 is to be formed using the numerals 0,1,2,3,4 and 5 without repetition.The total number of ways this can be done is
$\begin{array}{1 1}(A)\;216\\(B)\;240\\(C)\;600\\(D)\;3125\end{array} $
cbse
math
class11
ch7
permutations-and-combinations
additionalproblem
q17
sec-b
medium
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asked
May 20, 2014
by
sreemathi.v
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1 Answer
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$n!=n(n-1)(n-2)(n-3)......(3)(2)(1)$
Numbers using 1,2,3,4,5
No. of numbers =5!=120
Numbers using 0,1,2,4,5
No. of numbers =5!-4!=120-24=96
There are 4! numbers beginning with 0
$\therefore$ Total number of numbers =120+96=216
Hence (A) is the correct answer.
answered
May 20, 2014
by
sreemathi.v
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