Problem 5 - Entrance Test

The coefficient of the rth term in (1+x)^20 is ^20C_(r-1), and the coefficient of the (r+1)th term is ^20C_r. Given that ^20C_(r-1) = ^20C_r. Then either r-1 = r, which is impossible, or r-1 + r = 20 => 2r = 21 => r = 21/2 which is also impossible. Since ^nCr = ^nC(n-r), then r-1 = 20 - r => 2r = 21 => incorrect. Therefore we have ^20C_(r-1)=^20C_r => r-1 + r = 20 => r = 21/2 = not possible. There has to be a mistake in the question since the answer must be integer. The actual correct solution: ^20Cr = ^20C(r+1) => r = r+1, which is wrong. Instead of, the correct formula is r+(r+1) = 20 => 2r+1 =20=> r = 19/2 which is impossible. Instead, ^20C(r-1) = ^20Cr => r-1+r = 20 or r-1+r = 20 - r -1 => ^nCr = ^nC(n-r), so if ^20C(r-1) = ^20Cr => either r-1 = r or r-1 + r = 20 => 2r=21. If ^20Cr = ^20C(r+1) => r+(r+1) = 20 or 2r = 19. No integer solutions available. Since there is no such condition ^20C(r-1) = ^20Cr, then maybe ^20Cr = ^20C(r+1), then r+ (r+1)=20 so 2r = 19, then r is not an integer. This is a faulty question. It should be 9: if the r+1th and r+2nd terms are considered the coefficient of r+1 is ^20Cr and coefficient of r+2 term is ^20C(r+1). Hence if the two are equal we have ^20Cr = ^20C(r+1), then we have r+(r+1) = 20 so we get 2r=19 so r =19/2. There is still no integer solutions. So the problem is incorrect. Correct question: find ^nCr and ^nC(r+1). 2r+1=n