Question

Find the number of ways in which 10 doctor and 90 engineers can sit in a row having 100 chairs such that no doctor sit at either end of the row and between any two doctors, at least five engineers sit.

• A. $10!\times 90!$
• B. ${}^{100}{C}_{43}\times 10!\times 90!$
• C. ${}^{53}{C}_{43}\times 10!\times 90!$
• D. ${}^{90}{C}_{43}\times 10!\times 90!$

Answer 1

The correct option is C

${}^{53}{C}_{43}\times 10!\times 90!$

First we select 10 chairs which will be occupied by 10 doctors under the given condition. Now these 10

selected chairs will divide the remaining 90 chairs into 11 parts.

$x_1$ is the number of empty chairs on left side of first doctor, $x_{11}$ is the number of empty chairs on right side of

last doctor and $x_2,x_3,x_4............x_{10}$ are the number of empty chairs between any doctor (except the doctors who

sits on extreme right and extreme left)

No. of ways of selecting 10 chairs = No. of solution of

$x_1+x_2+x_3+x_4+...............+x_{11}=90$

Since, No doctor sits on either sides

So, chairs left on either sides should be greater than 1.

In between any two doctors, more than 5 engineers sit.

Conditions $x_1,x_{11}\ge 1,x_2,X_3,x_4..........+x_{10}\ge 5$

Coefficient of $x^{90}$ in $(x+x^2+x^3+...............{)}^2(x^5+x^6+................{)}^9$

Coefficient of $x^{43}$ in $(1+x+x^2+x^3+................{)}^{11}$

Coefficient of $x^{43}$ in $(1-x{)}^{-11}$ ${}^{53}{C}_{43}$

Hence required number of ways = ${}^{53}{C}_{43}$ × 10 ! × 90!