28. Necklaces You are putting nine blue glass beads, three red...

Question

28. Necklaces You are putting nine blue glass beads, three red glass beads, and seven green glass beads on a necklace. In how many distinguishable ways can the colored beads be put on the necklace?

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A box contains 5 red balls, 3 blue balls, and 2 green balls. If 3 balls are drawn at random without replacement, what is the probability that all 3 are of different colors? Awesome. So it appears, for this particular problem, we're asked to determine that if 3 balls are drawn at random without replacement. We're trying to determine what is the probability that all three of these particular balls that are drawn will be of different colors, so none of the colors will repeat. So what is the probability that all three of these 3 random balls drawn will be of different colors? So with that in mind, now that we know what we're ultimately trying to solve for, let's read off our multiple choice answers to see what our final answer might be. A is 1/8. B is 1/4. C is 1/2 and D is 1/16. Fantastic. So, first off, in order to solve this particular problem, we need to solve for the total number of balls. So we need to note that there's 5 red balls plus 3 blue balls, plus 2 green balls, which give us which will give us a total of 10 balls total. So I'll do a capital T for total. So now at this point we need to note once again that and also recall that we want to find the probability that 3 balls drawn without replacement are all of different colors, meaning we want 1 ball to be red, 1 ball to be blue, and 1 ball to be green. So with that in mind, our first step is we need to determine the number of favorable outcomes. So for 1 red out of 5, so if we were to get 1 red ball out of 5, that's a 5 to 1 chance, which will be equal to 5. And to get 1 blue out of 3, that's a 3 to 1 chance which will be equal to 3, and to get 1 green out of 2, that's a 2 to 1 chance which is equal to 2. So that means that our favorable outcomes, so I'll call this FO, so for favorable outcomes. We'll state 5 times 3 times 2, which will be equal to 30 total. So, as we should recall a note, since the order of drawing matters because we are drawing without replacement, and the question is about probability, we will need to count the number of favorable 3 ball sequences and not combinations. So, but, however, to be precise, We also need to calculate the total number of all possible ways to choose 3 balls from 10, regardless of color. So our second step now. is we need to determine the total number of possible outcomes. So we have 10. Choose 3 or 10 above 3. is equal to 10 multiplied by 9, multiplied by 8, all divided by 3 multiplied by 2 multiplied by 1. So when we take 10 multiplied by 9, multiplied by 8, all divided by 3 multiplied by 2 multiplied by 1, we will find that this is all equal to 120. Fantastic. So now, our 3rd step. We need to determine the final probability, so we need to find the probability P of one. Red, which I'll call R for short, so we need to find one red. One blue, which I'll call B for short, and 1 green which I'll call G for short. So the probability of getting one red, 1 blue, 1 green is going to be equal to 30 divided by 120, which will be equal to, when we plug it into our calculator, 1/4, and that is our final answer. Hooray, we did it. So we have a 1 to 4 chance of of selecting 3 balls at random that are all of different colors. And this corresponds with multiple choice answer letter B, which once again is 1/4. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

28. Necklaces You are putting nine blue glass beads, three red glass beads, and seven green glass beads on a necklace. In how many distinguishable ways can the colored beads be put on the necklace?

Key Concepts

Permutations and Combinations

Multiset Permutations

Circular Permutations

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