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

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: