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Question

# There are m seats in the first row of a theatre of which n are to be occupied. Number of ways of arranging n persons so that there should be at least two empty seats between any two persons

A
m2n+2Cnn!
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B
mn+1Cn (n - 2)!
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C
mn1Pn
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D
mnPnn!
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Solution

## The correct option is B m−2n+2Cnn!Given that n out of m seats are to be occupied. Since there are n people there are (n+1) spaces between them (including the starting and ending seats which may or may not be occupied).Let x1,x2,x3,......,xn+1 represent the spaces such that x1,xn+1≥0 ξ xi≥2Where i∈[2,n] and ⇒x1+x2+x3+......+xn+1+n=m⇒ Let x1+x=y1, xn+1=y and xi=yi−2 for i=2 to n.⇒yi≥0 for i=1 to n+1 and y1+(y2−2)+(y3−2)+.....+(yn−2)+yn+1=m−n⇒y1+y2+....+yn+1=m−n+2(n−1)⇒y1+y2+....+yn+1=m−n−2,yi≥0Total combinations is given by n+r−1Cr−1here, N=m+n−2,r=n+1Total =m+n−2+(n−1)−1C(n+1)−1 =m+2n−2CnNow, these n people can be arranged in their seats by n! ways.Total arrangement =n!×m+2n−2CnHence, the answer is n!×m+2n−2Cn.

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