`[(^n C_0+^n C_3+)1//2(^n C_1+^n C_2+^n C_4+^n C_5]^2+3//4(^n C_1-^n C_2+^n C_4-^n C_5+)^2=` `3` b. `4` c. `2` d. `1`

Prove that ^n C_0 ^(2n)C_n- ^n C_1 ^(2n-2)C_n+ ^n C_2^(2n-4)C_n-=2^ndot

Prove that ^n C_0^n C_0-^(n+1)C_1^n C_1+^(n+2)C_2^n C_2-=(-1)^ndot

Find the sum 3^n C_0-8^n C_1+13^n C_2 - 18^n C_3+..

Prove that "^n C_0^(2n)C_n-^n C_1^(2n-1)C_n+^n C_2xx^(2n-2)C_n++(-1)^n^n C_n^n C_n=1.

Using binomial theorem (without using the formula for ^n C_r ) , prove that "^n C_4+^m C_2-^m C_1^n C_2 = ^m C_4-^(m+n)C_1^m C_3+^(m+n)C_2^m C_2-^(m+n)C_3^m C_1+^ (m+n)C_4dot

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

Prove that ^n C_1(^n C_2)(^n C_3)^3(^n C_n)^nlt=((2^n)/(n+1))^(n+1_C()_2),AAn in Ndot

Prove that ^m C_1^n C_m-^m C_2^(2n)C_m+^m C_3^(3n)C_m-=(-1)^(m-1)n^mdot

Prove that (r+1)^n C_r-r^n C_r+(r-1)^n C_2-^n C_3++(-1)^r^n C_r = (-1)^r^(n-2)C_rdot

The value of determinant |[ ^n C_(r-1), ^n C_r, (r+1)^(n+2)C_(r+1)],[ ^n C_r, ^n C_(r+1),(r+2)^(n+2)C_(r+2)],[ ^n C_(r+1), ^n C_(r+2), (r+3)^(n+2)C_(r+3)]| is n^2+n-2 b. 0 c. ^n+3C_(r+3) d. ^n C_(r-1)+^n C_r+^n C_(r+1)