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A five digit number divisible by 3 is to be formed using the numericals 0,1,2,3,4 and 5 without repetition. If the total number of ways in which this can be done is n^3, then n ! must be

Option: 1

216


Option: 2

320


Option: 3

412

 


Option: 4

720


Answers (1)

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Number of numbers when 1,2,3,4,5 are used 5 !=120 and number of numbers when 0,1,2,4,5 are used

\begin{aligned} & =5 !-4 ! \\ \\& =96 \end{aligned}

\therefore Required number =120+96=216=n^3 \text { (given) }

\begin{array}{ll} \Rightarrow & n=6 \\ \\\text { then } & n !=6 !=720 \end{array}

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jitender.kumar

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