A bowl contains 9 red balls and 6 blue balls. In how many ways can 3 balls be selected so that both red and blue balls are obatined?

Question

I am stuck trying to figure out how to answer questions that ask how many ways more than one item can be selected from a number of items.

Answer 1

you have 9 red and 6 blue for a total of 15 balls

how many ways can you choose 3 balls out of 15 ball ?

15!/3!12!=455 ways to choose 3 balls

the only conditions you don't want are 3 red or 3 blue

how many ways can you choose 3 red balls from 9 red balls ?

9!/3!6!=84 ways to choose 3 red balls

how many ways can you choose 3 blue balls from 6 blue balls ?

6!/3!3!=20 ways to choose 3 blue balls

you have 84+20=104 conditions you don't want

455-104=351 ways to get either 2 red and 1 blue or 2 blue and 1 red

Answer 2

Hi Clayton,

 

Let's begin by determining the two scenarios in which the 3 selected balls from a bowl of only red and blue balls will contain both red balls and blue balls:

    1.  2 red balls and 1 blue ball.

    2.  2 blue balls and 1 red ball.

 

Now we'll determine the possible combinations for each of these two scenarios and then add together the results to obtain our answer:

    1.  Combinations for 2 red balls and 1 blue ball:

             (₉C₂)(₆C₁) = (36)(6) = 216

    2. Combinations for 2 blue balls and 1 red ball:

             (₆C₂)(₉C₁) = (15)(9) = 135

    3.  Total Combinations = 216 + 135 = 351

 

The key here was to first determine the possible scenarios and then compute the possible combinations for each one and finally combine the computed results for our answer.

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan.