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Question: There are \[m\] seats in the first row of a theatre, of which \[n\] are to be occupied. The number o...

There are mm seats in the first row of a theatre, of which nn are to be occupied. The number of ways of arranging nn persons so that there should be at least 2 empty seats between any two persons
A.m−2n+2Cnn!^{m - 2n + 2}{C_n}n!
B.m−n+1Cn(n−2)!^{m - n + 1}{C_n}\left( {n - 2} \right)!
C.m−n+1Pn^{m - n + 1}{P_n}
D.m−nPnn!^{m - n}{P_n}n!

Explanation

Solution

The number of seats mm is separated into number of people nn and number of empty places, say x.x., and we will be using this fact to solve for the number of arrangements of the people. A point to remember is that since there are nn people, the range goes from 1 to n+1n + 1 because we have to consider the starting and ending seats which may or may not be occupied.

Formula Used:
Total number of combinations = N+r−1Cr−1^{N + r - 1}{C_{r - 1}} …(i)

Complete step-by-step answer:
Given, the total number of seats = mm
The total number of seats occupied = nn
Let the number of spaces be xx

Let x1,x2,x3,...,xn+1{x_1},{x_2},{x_3},...,{x_{n + 1}} represent the spaces. So, it means,
x1+x2+x3+...+xn=x{x_1} + {x_2} + {x_3} + ... + {x_n} = x
And we have already established that
x+n=mx + n = m
Thus, x1+x2+x3+...+xn+n=m{x_1} + {x_2} + {x_3} + ... + {x_n} + n = m …(ii)
Let x1=y1,xn+1=yn+1{x_1} = {y_1},{\rm{ }}{x_{n + 1}} = {y_{n + 1}} and xi=yi−2,i∈[2,n]{x_i} = {y_i} - 2,{\rm{ }}i \in \left[ {2,n} \right]
We subtracted 2 from all the seats (except the first and last seat) so as to remove the possible spaces between them as the number of spaces between them is given to be 2, but we did not remove the seats from the first and last seats because they are on the ends and do not have the possibility of spaces.
So, doing the substitution of the seats and spaces into equation (ii) and we have got,
y1+(y2−2)+(y3−2)+(yn−2)+yn+1+n=m{y_1} + \left( {{y_2} - 2} \right) + \left( {{y_3} - 2} \right) + \left( {{y_n} - 2} \right) + {y_{n + 1}} + n = m
The number of yy terms with a −2 - 2 with it is - n−2+1=n−1n - 2 + 1 = n - 1 (because we are starting from y2{y_2})
Thus, we can club all the −2 - 2 terms together, and we get:
⇒\Rightarrow y1+y2+y3+...+yn+yn+1−2(n−1)+n=m{y_1} + {y_2} + {y_3} + ... + {y_n} + {y_{n + 1}} - 2\left( {n - 1} \right) + n = m
Now, taking 2(n−1),n2\left( {n - 1} \right),n to the other side, we get:
⇒\Rightarrow y1+y2+y3+...+yn+yn+1=m−n+2(n−1){y_1} + {y_2} + {y_3} + ... + {y_n} + {y_{n + 1}} = m - n + 2\left( {n - 1} \right)
Solving the right-hand side of the equation, we get:
⇒\Rightarrow y1+y2+y3+...+yn+yn+1=m−n+2n−2{y_1} + {y_2} + {y_3} + ... + {y_n} + {y_{n + 1}} = m - n + 2n - 2
⇒\Rightarrow y1+y2+y3+...+yn+yn+1=m+n−2{y_1} + {y_2} + {y_3} + ... + {y_n} + {y_{n + 1}} = m + n - 2
Now, with the total sum of all the yy terms we get the NN used in equation (i) and rr is the total number of units to be filled, which is n+1n + 1.
So, N=m+n−2N = m + n - 2 and r=n+1r = n + 1
Now, finally putting these values in equation (i) we get:
Total number of combinations, T=N+r−1Cr−1T{ = ^{N + r - 1}}{C_{r - 1}}
⇒\Rightarrow T=m+n−2+(n+1)−1Cn+1−1T{ = ^{m + n - 2 + \left( {n + 1} \right) - 1}}{C_{n + 1 - 1}}
⇒\Rightarrow T=m+2n−2CnT{ = ^{m + 2n - 2}}{C_n}
Now, these nn people can be arranged among themselves in n!n! ways.
Hence, the number of ways of arranging nn persons so that there should be at least 2 empty seats between any two persons is option A m+2n−2Cnn!^{m + 2n - 2}{C_n}n!.

Note: In these questions, we must remember that the seats on the ends do not have any places, so they won’t be considered in the transformation. Also, special tidiness must be ensured in solving such equations as there are a lot of terms and things get messy, increasing the chances of errors.