Question

There are 10 seats in the first row of a theatre of which 4 are to be occupied. The number of ways of arranging 4 persons so that no two persons sit side by side is
[a] ${}^7{C}_4$
[b] $4{\cdot }^7{P}_3$
[c] ${}^7{C}_3\cdot 4!$
[d] 840

Answer 1

Hint: Assume that a person has a chair with himself on which he sits. Hence keep only 6 seats in the row of the theatre. Now when a person sits, he chooses a gap between two seats and places his seat in between and in this way is seated in the theatre. Observe that in this way the persons will be separated from each other by at least one seat. Claim that the process of selection of 4 seats so that no two seats are consecutive is equivalent to the above gap selection and placement of seat process. Hence find the number of ways in which the persons can be seated so that no two persons sit side by side.

Complete step-by-step answer:
Let us name the seats on which the 4 persons do not sit as 1,2,…,6 and the seats on which the persons sit as A, B, C and D. Now, when the persons are seated so that no two persons sit side by side, the seats labelled as A, B, C and D have at least one seat among the seats labelled as 1,2,…,6 in between them. Hence the process of arranging the persons on the seats so that no two persons sit side by side is equivalent to the process of arranging seats labelled as 1,2,…,6 and the seats labelled as A, B, C and D in such a way that no two seats among the set A, B, C and D are adjacent taking care of the fact that the seats labelled 1 through 6 are identical(Since empty seats are indistinguishable).
We use gap methods to find the number of arrangements of the above condition.
Arrangement of 6 identical seats labelled 1 through 6 can be done in 1 way.
Now among the 6 seats, there are 7 gaps(denoted as Xs)
X 1 X 2 X 3 X 4 X 5 X 6 X .
Among the 7 gaps, four can be selected in ${}^7{C}_4$ ways
Now the arrangement of the seats A, B, C and D in these 4 places can be done in $4!$ ways.
Hence by the fundamental principle of counting the total number of ways in which the seats can be placed so that the above-mentioned condition is met is ${}^7{C}_{4}4!{=}^7{C}_{3}4!$
Hence option [c] is correct.

Note: Alternative Solution:
Let there be $x_1$ seats to the left of the first person, $x_2$ seats after that sits the next person, $x_3$ seats after that sits the third person, $x_4$ seats after that sits the fourth person and there are $x_5$ seats to the right of the last person
$x_1{P}_1{x}_2{P}_2{x}_3{P}_3{x}_4{P}_4{x}_5$
Hence, we have
$x_1+x_2+x_3+x_4+x_5=6,x_1\ge 0,x_2\ge 1,x_3\ge 1,x_4\ge 1,x_5\ge 0$
Let $y_1=x_1,y_2=x_2-1,y_3=x_3-1,y_4=x_4-1,y_5=x_5$
Hence, we have
$y_1,y_2,y_3,y_4,y_5\ge 0$ and $y_1+y_2+1+y_3+1+y_4+1+y_5=6\Rightarrow y_1+y_2+y_3+y_4+y_5=3$
We know that the number of non-negative integral solutions of the equation $x_1+x_2+x_3+\cdots +x_r=n$ is ${}^{n+r-1}{C}_{r-1}$
Hence, we have
The number of ways of choosing 4 seats so that no two chosen seats are consecutive is equal to the number of non-negative integral solutions of the equation $y_1+y_2+y_3+y_4+y_5=3$, which is equal to ${}^{3+5-1}{C}_{5-1}{=}^7{C}_4$
Among these 4 seats, 4 persons can be placed in 4! ways.
Hence the total number of ways ${=}^7{C}_{4}4!{=}^7{C}_{3}4!$, which is the same as obtained above.
Hence option [c] is correct.