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In a n- sided regular polygon, the proba...

In a `n-` sided regular polygon, the probability that the two diagonal chosen at random will intersect inside the polygon is: (a.)`(2^n C_2)/(^(^(n C_(2-n)))C_2)` (b.) `("^(n(n-1))C_2)/(^(^(n C_(2-n)))C_2)` (c.) `(^n C_4)/(^(^(n C_(2-n)))C_2)` (d.) none of these

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In a n- sided regular polygon, the probability that the two diagonal chosen at random will intersect inside the polygon is (2^n C_2)/(^(^(n C_(2-n)))C_2) b. (^(n(n-1))C_2)/(^(^(n C_(2-n)))C_2) c. (^n C_4)/(^(^(n C_(2-n)))C_2) d. none of these

In a n- sided regular polygon, the probability that the two diagonal chosen at random will intersect inside the polygon is (2^n C_2)/(^(^(n C_(2-n)))C_2) b. (^(n(n-1))C_2)/(^(^(n C_(2-n)))C_2) c. (^n C_4)/(^(^(n C_(2-n)))C_2) d. none of these

The value of (.^(n)C_(0))/(n)+(.^(n)C_(1))/(n+1)+(.^(n)C_(2))/(n+2)+"..."+(.^(n)C_(n))/(2n)

(.^(n)C_(0))^(2)+(.^(n)C_(1))^(2)+(.^(n)C_(2))^(2)+ . . .+(.^(n)C_(n))^(2) equals

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

C_(0)^(n)C_(n)^(n+1)+C_(1)^(n)C_(n-1)^(n)+C_(2)^(n)*C_(n-2)^(n-1)+.........+C_(n)^(n)*C_(0)^(1)=2^(n-1)(n+2)

Prove that ^nC_(0)^(2n)C_(n)-^(n)C_(1)^(2n-1)C_(n)+^(n)C_(2)xx^(2n-2)C_(n)++(-1)^(n)sim nC_(n)^(n)C_(n)=1