Permutations and Combinations · Question 33

Let nCr-1 = 28, nCr = 56 and nCr+1 = 70. Let A(4 t, 4 t), B(2 t, -2 t) and C (3r - n, r2 - n - 1 ) be the vertices of a triangle ABC, where t is a parameter. If (3x - 1)2 + (3y)2 = , is the locus of the centroid of triangle ABC, then equals

1. 18
2. 8
3. 20
4. 6

Answer

(3) 20

Solution

nCrnCr-1 = n - r + 1r = 5628 = 2 n + 1 = 3r

nCr+1nCr = n - rr + 1 = 7056 = 54 4n = 9r + 5

4(3r - 1) = 9r + 5 r = 3, n = 8

C = (9 - 8, 9 - 8 - 1) = (1, 0)

Centroid (x, y):

3x = 4 t + 2 t + 1 3x - 1 = 4 t + 2 t

3y = 4 t - 2 t

Squaring and adding:

(3x - 1)2 + (3y)2 = 16( 2 t + 2 t) + 4( 2 t + 2 t) + 16 t t - 16 t t

= 20

= 20

Answer: (3)