Question

In how many ways can $12$ gentlemen sit around a round table so that three specified gentlemen are always together?

• A. 9!
• B. 10!
• C. 3! 10!
• D. 3! 9!

Hint:

 in the space for \(3\) gentlemen we can permute them in \(3 !\)

Answer 1

The Correct Option is D

Let us consider the \(3\) specific gentlemen as a single person that occupies three times the space. 
So, now we have \(10\) persons who can be made to sit in \(9 !\) ways and in the space for \(3\) gentlemen we can permute them in \(3 !\) ways. 
So, total ways are \(3 !\, 9 !\).

Discover More Topics From This Chapter: Permutations and Combinations

Answer 2

The Correct Answer is (D)

Real Life Applications

• This issue could occur in the real world if the team lead's four members are required to constantly sit together. They might be the four board members with the most experience, or they could be the members in charge of various areas of business operations. To facilitate collaboration and communication, it would be crucial to ensure that they are constantly sat together in this situation. 
• This issue can also come up in a courtroom or jury situation. It would be crucial to ensure that five jurors are always seated together if they are known to be biassed in favour of one side of a case in order to prevent their bias from influencing the verdict.

Question can also be asked as

• In how many ways can 12 gentlemen be arranged around a round table if 3 of them must sit together? 
• How many ways are there to seat 12 gentlemen around a circular table if 3 of them must be seated together? 
• The 12 gentlemen are to be seated around a circular table. 3 of them must sit together. How many seating arrangements are there?

Answer 3

The Correct Answer is (D)

Combination and Permutation In all facets of real-world counting situations, formulas are applied. 

• Permutations Formula is given as P(n,r) = n!/(n-r)!, where [n>= r].
• Combinations Formula is given as C(n,r) = n!/[r !(n-r)!], where [n>= r].

Derivation of Permutations Formula

P (n, r) = n . (n-1) . (n-2) . (n-3)…… (n-(r-1)) … Ways.

It can be written as

P (n, r) = n . (n-1) . (n-2) . (n-3) …. (n-r+1) …….. (1)

On multiplying and dividing (1) by (n-r) (n-r-1) (n-r-2)........... 3.2.1, 

\(P (n, r) =\frac{n.(n-1).(n-2).…. (n-r+1)[(n-r) (n-r-1) (n-r-2)... 3. 2. 1]}{[(n-r) (n-r-1) (n-r-2)....3. 2. 1]}\)

\(p (n, r)= \frac{n!}{(n-r)!}\)

Derivation of Combinations Formula

Combination is a method of selecting r items from a list of n items where the order is irrelevant. The number of ways to arrange r things in r ways equals r!, according to the basic counting concept.

\(C(n,r) =\dfrac{\frac{n!}{(n-r)!}}{r!}.\)

\(C(n, r) = \dfrac{n!}{r!.(n - r)!}\)

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Probability and Statistics	Important Questions For Class 11 Maths Permutations and Combinations	Factorial Formula