Question

Assume that you have 6 dimes and 2 quarters (all distinct), and you select 4 coins. (1) In how many ways can the selection be made? equation editorEquation Editor (2) In how many ways can the selection be made if all the coins are dimes? equation editorEquation Editor (3) In how many ways can the selection be made if you select 3 dimes and 1 quarter? equation editorEquation Editor (4) In how many ways can the selection be made so that at least 3 coins are dimes?

Answer 1

Answer:

Step-by-step explanation:

Given that there are 6 dimes and 2 quarters (all distinct), and you select 4 coins.

i) No of ways to select 4 coins = $(6+2)C4 = 70$

ii) No of ways to select 4 dimes = $6C4 = 15$

iii) No of ways to select 3 dimes and 1 qr = $6C3(2C1)\\= 20(2)\\=40$

iv) no of ways to select atleast 3 dimes = no of ways for 3 dimes + no of ways for 4 dimes

= $40+15 =55$