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A double-decker bus carry (u+e) passenge...

A double-decker bus carry `(u+e)` passengers, `u` in the upper deck and `e` in the lower deck. Find the number of ways in which the `u+e` passengers can be distributed in the two decks, if `r(lt=e)` particular passengers refuse to go in the upper deck and `s(lt=u)` refuse to sit in the lower deck.

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The correct Answer is:
`((u+l-r-s)!)/((l-r)!(u-s)!)`
Number of passengers to be allotted in the two decks bus
=(u+1)-(r+s)
Number of seats available in the upper deck
=u(u-s)
Number of seats available in lower deck =l-r
`therefore` Total number of ways of distribution `=((u+l-r-s)!)/((l-r)!(u-s)!)`
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Let A be an mxxn matrix. If there exists a matrix L of type nxxm such that LA=I_(n) , then L is called left inverse of A. Similarly, if there exists a matrix R of type nxxm such that AR=I_(m) , then R is called right inverse of A. For example, to find right inverse of matrix A=[(1,-1),(1,1),(2,3)] , we take R=[(x,y,x),(u,v,w)] and solve AR=I_(3) , i.e., [(1,-1),(1,1),(2,3)][(x,y,z),(u,v,w)]=[(1,0,0),(0,1,0),(0,0,1)] {:(implies,x-u=1,y-v=0,z-w=0),(,x+u=0,y+v=1,z+w=0),(,2x+3u=0,2y+3v=0,2z+3w=1):} As this system of equations is inconsistent, we say there is no right inverse for matrix A. For which of the following matrices, the number of left inverses is greater than the number of right inverses ?