Question

In how many ways \[10\] persons \[{A_1},{A_2},{A_3},{A_4},...,{A_{10}}\] can be seated along a row such that
i) \[{A_1},{A_2},{A_3}\] sit together
ii) \[{A_1},{A_2},{A_3}\] sit in a specified order need not be together
iii) \[{A_1},{A_2},{A_3}\] sit together in a specified order
 iv) \[{A_2},{A_3},{A_4}\] sit always after \[{A_1}\].

Answer 1

Hint: First we know a permutation is an act of arranging the objects or numbers in order. Consider each option, using the formula \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\;\;}}\] where \[n\] is the total items in the set and \[r\] is the number of items taken for the permutation to find the value of \[n\].

Complete step by step solution:
The factorial of a natural number is a number multiplied by "number minus one", then by "number minus two", and so on till \[1\] i.e., \[n! = n \times \left( {n - 1} \right) \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times - - - \times 2 \times 1\]. The factorial of \[n\] is denoted as \[n!\].
Given \[10\] persons \[{A_1},{A_2},{A_3},{A_4},...,{A_{10}}\].

(i) Given \[{A_1},{A_2},{A_3}\] sit together.
Then \[{A_1},{A_2},{A_3}\] are taken together, we may treat them as one person. Then, the total number of persons becomes \[8\].
The number of ways \[8\] persons can be seated along a row is \[{}^8{P_8} = 8! = 40320\].
The persons \[{A_1},{A_2},{A_3}\] may be arranged in \[{}^3{P_3} = 3! = 6\] ways.
Hence, the total number of ways where \[{A_1},{A_2},{A_3}\] are sit together \[ = 40320 \times 6 = 2,41,920\] ways.

(ii) Given\[{A_1},{A_2},{A_3}\] sit in a specified order need not be together.
Total number of ways where the \[10\] persons can be seated along a row \[10! = 36,28,800\]ways.
Since from the option (i), the total number of ways where \[{A_1},{A_2},{A_3}\] are sit together \[ = 2,41,920\] ways.
Hence, total number of ways \[{A_1},{A_2},{A_3}\] sit in a specified order need not be together \[ = 36,28,800 - 2,41,920 = 33,86,880\] ways.

(iii) Given \[{A_1},{A_2},{A_3}\] sit together in a specified order
Suppose the persons sit in the \[{A_1},{A_2},{A_3}\]
Then the total number of ways where \[{A_1},{A_2},{A_3}\] sit together in a specified order=\[40320\]ways.

(vi) Given \[{A_2},{A_3},{A_4}\]sit always after \[{A_1}\].
Total number of ways if \[{A_2},{A_3},{A_4}\] sit always after \[{A_1}\]\[ = 9! + 6 \times 8! + 15 \times 7! + 20 \times 6! + 15 \times 5! + 6 \times 4! + 3! = 6,96,750\] ways.

Note: