Question

Consider a bag containing four red marbles, three green ones, one transparent one, three yellow ones, and three orange ones.

Answer 1

Answer:

21

Step-by-step explanation:

Given,

• Number of red marble = 4
• number of green marble = 3
• number of transparent marble = 1
• number of yellow marble = 3
• number of orange marble = 3

Total number of marble except red and green  = 3 +3+1

                                                                               = 7

So, the total number of possible sets of five marbles such that none of them are green or red can be given by

$n\ =\ ^7C_5$

   $=\ \dfrac{7!}{(7-5)!.5!}$

   $=\ \dfrac{7!}{5!.2!}$

  $=\ \dfrac{42}{2}$

   = 21

So, the required number of possible sets are 21.