Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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 <img alt="n^3" src="https:/

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)

best_answer

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}

Posted by

jitender.kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question