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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!×90!

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B

100C43×10!×90!

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C

53C43×10!×90!

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D

90C43×10!×90!

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Solution

The correct option is C

53C43×10!×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.

x1 is the number of empty chairs on left side of first doctor, x11 is the number of empty chairs on right side of

last doctor and x2,x3,x4............x10 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

x1+x2+x3+x4+...............+x11=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 x1,x111,x2,X3,x4..........+x105

Coefficient of x90 in (x+x2+x3+...............)2(x5+x6+................)9

Coefficient of x43 in (1+x+x2+x3+................)11

Coefficient of x43 in (1x)11 53C43

Hence required number of ways = 53C43 × 10 ! × 90!


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