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A bag contains 3 red marbles 2 blue marbles and 4 green marbles_ Each color type has distinguishing feature_ If 3 marbles are chosen randomly; how many ways can you...

Question

A bag contains 3 red marbles 2 blue marbles and 4 green marbles_ Each color type has distinguishing feature_ If 3 marbles are chosen randomly; how many ways can you select at least 2 green marblesAnswer:

A bag contains 3 red marbles 2 blue marbles and 4 green marbles_ Each color type has distinguishing feature_ If 3 marbles are chosen randomly; how many ways can you select at least 2 green marbles Answer:



Answers

A bag contains three red marbles, two green ones, one lavender one, two yellows, and two orange marbles. How many possible sets of three marbles are there?

I had to set up this problem. We have three Red Marbles um too green. We have one lavender. Never heard of lavender in a math problem before. Two yellows. And then to orange. Alright. Oh are so it doesn't look like 20. Um But anyway if you can add all those up 3, 6789, 10 Marbles in total. And what really matters here is you know, let's say somebody were to choose all three red with one neat lavender. That would be the same as choosing the first lavender first and then there are other three red. So what I'm saying is the selection of how you choose them. Doesn't matter. um they would just want 10 combination for And so how we perform this arithmetic is 10 factorial over the second number of factorial and then you subtract between number's factorial. So the way I like to teach this problem is to just write out actually like doing stopping at six factorial but I'll go ahead and show all the way down, Multiplying down to one. four factorial is four times 3 attempts to times one and then six factorial six times five times four times three times two times one. That's why I would stop at six factorial because all of those pieces after six factorial would cancel anything divided by itself is one. Same thing with forward. About four times two is the same thing as eight. And then while I'm at it I would probably rewrite nine divided by three as three. So as I'm looking at this problem I would just do three times seven, is 21 times 10 is 210 and divided by that one would still be 210. So that should be your final answer. 210 different ways of choosing these four marbles.

Yeah. So here's problem. 52 I'm going to read it. Says Bag contains one green marble, one red marble and one blue marble. Um and then we've got a diagram that shows the possible outcomes of randomly drawing three marbles from the bag without replacement. And it says how Maney combinations of three marbles could be drawn from the bag. Um, so we'll answer that part first is about the combinations. Um, and the key thing here about a combination is that the order doesn't matter. Eso it doesn't matter if I pick the red marble first, the blue marble first the green marble First it's the same thing. Um, so if I want to draw three marbles from this group of three, um, there's only one way to do that because I'm picking all of the marbles and I don't care what order I'm sort of doing that in. So there's only one combination. Um, and sometimes we write this situation as there's a set of three marbles, and we are choosing three of them, so that values equal to 13 choose three is how you say that now, with permutations, Order matters so it makes a difference if I pick the green one first, then the red one first, then the blue one first. So I am picking three separate things. Sometimes I represent these with separate little slots and the first thing before I've picked anything and I'm not replacing. So I'm not putting the marble back in the first thing. When I pick it, I could pick the green one. I could pick the red one. I could pick the blue one to those three separate possibilities for what I could pick. Um, Then in the second slot here, I've picked one of the marbles. There's only two left. So two possibilities for the second slot and then the third slot. The third thing I pick there's only one possibility because I replaced didn't replace any of the marbles. So when the order matters, we could go through and use this process to figure out, um, how many different ways you could order these marbles, basically. And the answer there is three times, two times one is six. Um, so two very different answers, depending on whether I'm caring about the order or whether I'm not

This question, we are given a jar that contains 10 Red Marbles and 12 Blue Marbles, and we're basically asked how many ways there are to draw throwing red marbles and to will marbles. So the number of waves, basically, we Choose three from 10, and then we choose two from blue, So that gives us 7, 9, 20 different ways.

In this question it is given that we have eight marble pieces and out of those four is red too is grain and two is yellow. Now it is told that we have to select three marbles and the selection must be of the kind that there should not be all marble of the same color. So we can see that this can be done by the total selections minus those elections in which we have marble solved same color. Now if we look about total selection then out of it we have to select C. three. So this is done by C83, no marvel of same color meals, All the three are from red, So it is done by C43. No value of C8 threes 56 and very well C43 is four. And after solving it we get 52 So 52 is the required answer of this question here. Thank you.


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