QA – Placement Quizzes | Permutation and Combination | Question 10

A box contains 2 red coins, 3 green coins and 4 blue coins. In how many ways can 3 coins be chosen such that at least one coin is green?
(A) 16
(B) 32
(C) 64
(D) 128


Answer: (C)

Explanation: There are three cases:

1. 3 green coins
2. 2 green coins + 1 non-green coin
3. 1 green coin + 2 non-green coins

Therefore, total number of ways = ³C₃ + ³C₂ * ⁶C₁ + ³C₁ * ⁶C₂ = 1 + 3*6 + 3*15 = 64.

Quiz of this Question
Please comment below if you find anything wrong in the above post