Instruction: You must show all of your work bysl setting up the problem using combinatorics and simplifying completely.

question: There are 5 red marbles and 9 blue marbles in a bag. suppose two of the marbles are cracked. how many ways can four marbles be selected so that either no cracked marbles are selected or just one cracked marble is selected?

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Mathematics

7 September, 01:23

Instruction: You must show all of your work bysl setting up the problem using combinatorics and simplifying completely.

question: There are 5 red marbles and 9 blue marbles in a bag. suppose two of the marbles are cracked. how many ways can four marbles be selected so that either no cracked marbles are selected or just one cracked marble is selected?

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Cisco · Answer 1

There are 14 marbles in the bag, and we either want to get 0 or 1 cracked one (out of 2 cracked marbles) in 4 draws. Since there are 12 uncracked marbles and 2 cracked ones:

To get 0 cracked marbles, there are 12C4 ways (to draw 4 marbles out of the 12 uncracked ones only). 12C4 = 495 ways

To get 1 cracked marble, we need 3 to be from the 12 uncracked ones, and 1 from the 2 cracked ones. This means there are 12C3 x 2C1 = 440 ways

So the total number of ways is 495 + 440 = 935 ways.