A man wants to distribute 101 coins of a rupee each, among his 3 sons with the condition that no one receives more money than the combined total of the other two. The number of ways of doing this is equal to

Step-by-step solution

Let, the amount received by the sons be Rs. x , Rs. y , and Rs z respectively, then x ≤ y + z = 101 - x ⇒ 2 x ≤ 101 . ⇒ x ≤ 50 , y ≤ 50 , z ≤ 50 . x + y + z = 10 1 So, the required number of ways is the coefficient of x 101 in 1 + x + x 2 + . . . . + x 51 3 or coefficient of x 101 in the expansion of 1 - x 50 1 - x 3 or coefficient of x 101 in the expansion of 1 - x 51 3 1 - x - 3 or coefficient of x 101 in the expansion of 1 - 3 C 1 x 51 + . . . . . 1 - x - 3 or coefficient of x 101 in the expansion of 1 - x - 3 -