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n points are given of which r points are...

n points are given of which r points are collinear, then the number of straight lines that can be found = (a) `""^(n)C_(2)-""^(r)C_(2)` (b) `""^(n)C_(2)-""^(r)C_(2)+1` (c) `""^(n)C_(2)-""^(r)C_(2)-1` (d) None of these

A

`""^(n)C_(2)-""^(r)C_(2)`

B

`""^(n)C_(2)-""^(r)C_(2)+1`

C

`""^(n)C_(2)-""^(r)C_(2)-1`

D

None of these

Text Solution

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The correct Answer is:
B
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""^(n) C_(r+1)+2""^(n)C_(r) +""^(n)C_(r-1)=

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""^(n)C_(n-r)+3.""^(n)C_(n-r+1)+3.""^(n)C_(n-r+2)+""^(n)C_(n-r+3)=""^(x)C_(r)

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""^(n-2)C_(r)+2""^(n-2)C_(r-1)+""^(n-2)C_(r-2) equals :

""^(n)C_(r)+2""^(n)C_(r-1)+^(n)C_(r-2) is equal to

The expression ""^(n)C_(r)+4.""^(n)C_(r-1)+6.""^(n)C_(r-2)+4.""^(n)C_(r-3)+""^(n)C_(r-4)

If n is even and ""^(n)C_(0)lt""^(n)C_(1) lt ""^(n)C_(2) lt ....lt ""^(n)C_(r) gt ""^(n)C_(r+1) gt""^(n)C_(r+2) gt......gt""^(n)C_(n) , then, r=

f(n)=sum_(r=1)^(n) [r^(2)(""^(n)C_(r)-""^(n)C_(r-1))+(2r+1)(""^(n)C_(r ))] , then