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Chapter 6: Problem 40

A bag contains three red marbles, two green ones, one lavender one, two yellows, and two orange marbles. How many sets of five marbles include at most one of the red ones?

Short Answer

Expert verified
There are 126 sets of five marbles that include at most one of the red marbles.

Step by step solution

01

Scenario 1: No Red Marbles

If we have no red marbles in the set, then we will take marbles from the other colors only. There are 7 non-red marbles (2 green, 1 lavender, 2 yellow, and 2 orange), and we want to choose 5 of them. The number of combinations in this scenario can be found using the combination formula: C(n, r) = \(\frac{n!}{r!(n-r)!}\) Here, n is the total number of marbles, and r is the number of marbles we want to choose. So, C(7, 5) = \(\frac{7!}{5!(7-5)!}\) = \(\frac{7!}{5!2!}\)
02

Calculate the Combinations in Scenario 1

Now let's calculate the combinations: C(7, 5) = \(\frac{7!}{(5!)(2!)}\) = \(\frac{7*6*5*4*3*2*1}{(5*4*3*2*1)(2*1)}\) = \(\frac{7*6}{2}\) = 21 So there are 21 sets with no red marbles.
03

Scenario 2: One Red Marble

Now let's consider the case where there is exactly one red marble in the set. In this scenario, we need to choose 1 red marble from the 3 red marbles, and 4 other marbles from the remaining 7 non-red marbles (2 green, 1 lavender, 2 yellow, and 2 orange). The number of combinations for choosing one red marble is: C(3, 1) = \(\frac{3!}{1!(3-1)!}\) = \(\frac{3!}{1!2!}\) The number of combinations for choosing four non-red marbles from the remaining seven is: C(7, 4) = \(\frac{7!}{4!(7-4)!}\) = \(\frac{7!}{4!3!}\)
04

Calculate the Combinations in Scenario 2

Let's calculate the number of combinations for each part of this scenario: For one red marble: C(3, 1) = \(\frac{3!}{(1!)(2!)}\) = \(\frac{3*2*1}{(1)(2*1)}\) = 3 For four non-red marbles: C(7, 4) = \(\frac{7!}{(4!)(3!)}\) = \(\frac{7*6*5*4*3*2*1}{(4*3*2*1)(3*2*1)}\) = \(\frac{7*6*5}{2*1}\) = 35 Now, to get the total number of combinations in this scenario, we multiply the combinations of each part together: Total combinations for one red marble and four non-red marbles = 3 * 35 = 105
05

Final Answer

Finally, we sum the combinations from both scenarios: Total sets of 5 marbles with at most one red one = Sets with no red marbles + Sets with one red marble = 21 + 105 = 126 So there are 126 sets of five marbles that include at most one of the red marbles.

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Most popular questions from this chapter

Let \(S\) be the set of outcomes when two distinguishable dice are rolled, let \(E\) be the subset of outcomes in which at least one die shows an even number, and let \(F\) be the subset of outcomes in which at least one die shows an odd number. List the elements in each subset given. $$ E^{\prime} \cup F^{\prime} $$

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