Your urn contains 8 red balls, and 9 blue balls, and you draw 4 balls. In how many ways can the selection be made so that at least one of each color is drawn?

Question

Your urn contains 8 red balls, and 9 blue balls, and you draw 4 balls.
In how many ways can the selection be made so that at least one of each color is drawn?

Answer 1

Hi Elizabeth,

 

Let's begin by determining the ways in which you would have at least a ball of each color with 4 balls drawn from an urn of 8 red balls and 9 blue balls:

    1.  3 red balls and 1 blue ball.

    2.  2 red balls and 2 blue balls.

    3.  1 red ball and 3 blue balls.

 

We see that there are three scenarios for drawing at least a ball of each color.  Therefore, we will calculate the possible combinations for each scenario and then take their sum for our answer:

    1.  3 red balls and 1 blue ball:

             (₈C₃)(₉C₁) = (56)(9) = 504

    2.  2 red balls and 2 blue balls:

             (₈C₂)(₉C₂) = (28)(36) = 1,008

    3.  1 red ball and 3 blue balls:

             (₈C₁)(₉C₃) = (8)(84) = 672

    Total Combinations = 504 + 1,008 + 672 = 2,184

 

The key to remember in these combination problems is to first identify each valid scenario and then compute the combinations for each scenario and then sum the results for the final answer.

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan.