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Class 11MathematicsPermutations and CombinationsPermutationsExpert
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(5) distinct white balls and (3) distinct black balls are placed around a circle. If no two black balls are adjacent, how many arrangements are possible?

Answer and explanation

Correct answer: (2880)

First arrange the (5) white balls around a circle in ((5-1)!) ways, then place (3) black balls in (^{5}P_{3}) circular gaps. Exam tip: in circular gaps, arrange the base objects first.

Related tags

PermutationsCircular GapsNo AdjacencyClass 11Permutations 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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