In a bag there are...

Question

In a bag there are 7 green, 11 brown, and 8 red marbles. In how many ways can 5 marbles be randomly selected from the bag, without replacement, if the selection shows all 5 marbles of the same color?

Answer 1

Hi Shelby,

 

Let's begin by seeing how many different scenarios there are for selecting 5 marbles of the same color without replacement:

    1.  Successively select 5 green marbles.

    2.  Successively select 5 brown marbles.

    3,  Successively select 5 red marbles.

 

What we need to do now is determine the following:

    1.  The number of possible ways to select a combination of 5 green marbles from the 7 green marbles in the bag.

    2.  The number of possible ways to select a combination of 5 brown marbles from the 11 brown marbles in the bag.

    3.  The number of possible ways to select a combination of 5 red marbles from the 8 red marbles in the bag.

    4.  The sum of ALL these three color combination results will be our answer.

 

Here then are our calculations:

    1.  Ways to choose 5 out of 7 green marbles:

             ₇C₅ = 7!/[(7-5)!(5!)]

             ₇C₅ = 7!/[(2!)(5!)] = 21

    2.  Ways to choose 5 out of 11 brown marbles:
             ₁₁C₅ = 11!/[(11-5)!(5!)]

             ₁₁C₅ = 11!/[(6!)(5!)] = 462

    3.  Ways to choose 5 out of 8 red marbles:

             ₈C₅ = (8!)/[(8-5)!(5!)]

             ₈C₅ = 8!/[(3!)(5!)] = 56

    4.  Total ways to choose 5 marbles of same color:

             ₇C₅ + ₁₁C₅ + ₈C₅ = 21 + 462 + 56 = 539

 

Thus, our answer is that there are 539 ways to choose 5 marbles of the same color without replacement.

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan.