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Class 11MathematicsPermutations and CombinationsPermutationsHard
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In how many ways can (8) distinct people be seated around a circular table so that two particular people do not sit together?

Answer and explanation

Correct answer: (3600)

Total circular arrangements are (7!), and together cases are (6!\cdot2!). Hence the answer is (5040-1440=3600).

Related tags

PermutationsCircular SeatingNot TogetherPermutations And CombinationsMathematicsClass 11 Mcq
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10 more MCQs
  1. 1A menu offers 5 types of snacks and 4 types of drinks. In how many ways can one snack and one drink be chosen?
  2. 2An online form has 4 options for class, 3 options for section, and 5 options for subject. In how many ways can one complete selection of a class, a section, and a subject be made?
  3. 3How many (3)-digit numbers can be formed from digits (1,3,5,7,9) without repetition?
  4. 4A (3)-letter code is to be formed from (6) distinct letters. If repetition is not allowed, how many codes are possible?
  5. 5Three distinct prizes are to be awarded to seven students. No student can receive more than one prize. How many ways are possible?
  6. 6If \(\frac{(n+5)!}{(n+3)!}=210\), what is the value of \(n\)?
  7. 7What is the value of \(\frac{21!}{19!\cdot2!}-\frac{18!}{16!\cdot2!}\)?
  8. 8If \(\frac{(n+1)!}{(n-1)!}-\frac{n!}{(n-2)!}=24\), what is the value of \(n\)?
  9. 9What is the value of \(\frac{22!}{20!\cdot 2!}-\frac{21!}{20!}\)?
  10. 10If (5) different books are arranged in a row, how many ways are possible?
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