M metru New member Joined Jul 22, 2005 Messages 3 Mar 9, 2006 #1 Prove: \(\displaystyle C_{k - 3}^{k - 3} \cdot C_{n - k + 2}^2 + C_{k - 2}^{k - 3} \cdot C_{n - k + 1}^2 + C_{k - 1}^{k - 3} \cdot C_{n - k}^2 + \cdots + C_{n - 3}^{k - 3} \cdot C_2^2 = C_n^k\)
Prove: \(\displaystyle C_{k - 3}^{k - 3} \cdot C_{n - k + 2}^2 + C_{k - 2}^{k - 3} \cdot C_{n - k + 1}^2 + C_{k - 1}^{k - 3} \cdot C_{n - k}^2 + \cdots + C_{n - 3}^{k - 3} \cdot C_2^2 = C_n^k\)
D Denis Senior Member Joined Feb 17, 2004 Messages 1,700 Mar 9, 2006 #2 I proved it using a napkin at Burger King; sorry, lost the napkin :roll:
J jolly Junior Member Joined Apr 18, 2004 Messages 124 Mar 9, 2006 #3 Denis said: I proved it using a napkin at Burger King; sorry, lost the napkin :roll: Click to expand... :lol:
Denis said: I proved it using a napkin at Burger King; sorry, lost the napkin :roll: Click to expand... :lol:
