Question

A bag contains 2 gold coins, 3 silver coins and 4 bronze coins. in how many ways can 3 coins be drawn from the bag, if at least one silver coin is to be included in the draw?

Answer 1

Answer:

Total number of ways = 64

Step-by-step explanation:

It is given that, A bag contains 2 gold coins, 3 silver coins and 4 bronze coins.

Therefore, there are 9 coins in the bag

The possible ways are

1)three coin are silver

2)2 coins are silver and one coin is either gold or bronze

3) only one coin is silver other coins are either gold or bronze

If three coins be silver

The possible ways = 3C₃ = 1

If 2 coins be silver

The possible ways = 3C₂x6C₁ = 18

If one coin is silver

The possible ways = 45

Therefore total number of possible ways = 1 + 18 + 45 =64 ways

Answer 2

Answer:

64 ways

Step-by-step explanation:

A bag contains 2 gold coins, 3 silver coins and 4 bronze coins.

We have to drawn 3 coins from the bag, if atleast one silver coin is to be include.

First we write possible way

Let G denote drawn gold coin

Let S denote drawn silver coin

Let B denote drawn bronze coin

Condition on silver coin

Total number of silver coin, 3

Total number of other coin(G+B), 6

Case 1: For 3 drawn, One silver and 2 other

$^3C_1\times ^6C_2$

Case 2: For 3 drawn, Two silver and 1 other

$^3C_2\times ^6C_1$

Case 3: For 3 drawn, 3 silver and 0 other

$^3C_3\times ^6C_0$

Total number of ways,  

$\Rightarrow ^3C_1\times ^6C_2+^3C_2\times ^6C_1+^3C_3\times ^6C_0$

$\Rightarrow 3\times 15+3\times 6+1\times 1$

$\Rightarrow 45+18+1$

$\Rightarrow 64$

Thus, The total number of ways can 3 coins be drawn from the bag, if atleast one silver coin is 64