If `alpha= ""^m C_2`, then `""^(alpha)C_2` is equal to
a. `""^(m+1)C_4`
b. `""^(m-1)C_4`
c. 3 `""^(m+2)C_4`
d. 3 `""^(m+1)C_4`

Solve the inequality .^(n-1)C_4 - .^(n -1)C_3 - 5/4 .^(n-2)C_2 lt 0,n in N

The value of expression .^47 C_4+sum_(j=1)^5.^(52-j)C_3 is equal to a. .^47 C_5 b. .^52 C_5 c. .^52 C_4 d. none of these

In triangle A B C , if r_1=2r_2=3r_3, then a : b is equal to 5/4 (b) 4/5 (c) 7/4 (d) 4/7

If m=.^(n)C_(2) show that , .^(m)C_(2)=3.^(n+1)C_(4)

If m= ^nC_2 show that ^mC_2=3^(n+1)C_4

C_0/2-C_1/3+C_2/4-C_3/5+.... is equal to

Using binomial theorem (without using the formula for .^n C_r ) , prove that .^nC_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

Let m in N and C_(r) = ""^(n)C_(r) , for 0 le r len Statement-1: (1)/(m!)C_(0) + (n)/((m +1)!) C_(1) + (n(n-1))/((m +2)!) C_(2) +… + (n(n-1)(n-2)….2.1)/((m+n)!) C_(n) = ((m + n + 1 )(m+n +2)…(m +2n))/((m +n)!) Statement-2: For r le 0 ""^(m)C_(r)""^(n)C_(0)+""^(m)C_(r-1)""^(n)C_(1) + ""^(m)C_(r-2) ""^(n)C_(2) +...+ ""^(m)C_(0)""^(n)C_(r) = ""^(m+n)C_(r) . (a) Statement-1 and Statement-2 both are correct and Statement-2 is the correct explanation for the Statement-1. (b) Statement-1 and Statement-2 both are correct and Statement-2 is not the correct explanation for the Statement-1. (c) Statement-1 is correct but Statement-2 is wrong. (d) Statement-2 is correct but Statement-1 is wrong.

The value of .^(n)C_(1)+.^(n+1)C_(2)+.^(n+2)C_(3)+"….."+.^(n+m-1)C_(m) is equal to

Let A D be a median of the A B Cdot If A Ea n dA F are medians of the triangle A B Da n dA D C , respectively, and A D=m_1,A E=m_2,A F=m_3, then (a^2)/8 is equal to (a) m_2^2+m_3^2-2m_1^2 (b) m_1^2+m_2^2-2m_3^2 (c) m_1^2+m_3^2-2m_2^2 (d) none of these