Question

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

How many possible sets of five marbles are there in which none of them are red or green?

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.