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The different faces of a cuboid, having the measurements X2X 2 Cm, is to be painted. Colors at disposal arC blue, yellow, and red. In how many ways can this be done...

Question

The different faces of a cuboid, having the measurements X2X 2 Cm, is to be painted. Colors at disposal arC blue, yellow, and red. In how many ways can this be done; if two colorings are considered as being equivalent when one of them can be obtained from the other one by a rotation of the cuboid?

The different faces of a cuboid, having the measurements X2X 2 Cm, is to be painted. Colors at disposal arC blue, yellow, and red. In how many ways can this be done; if two colorings are considered as being equivalent when one of them can be obtained from the other one by a rotation of the cuboid?



Answers

Each of the six squares shown in the figure is to be filled with any one of ten possible colors. How many ways are there of coloring the strip shown in the figure so that no two adjacent squares have the same color?

All right. So here are interested in that. Determining how many ways eh to you by an checkerboard can be tiled using one bite to or to write to pieces. So if we have a checkerboard is n and squares long, then we have a few different options. The first thing we could do is place a one by two piece vertically in the far right. And then we have a two by and minus one second boy remaining. And so if we assume that the number of ways this entire checkerboard could be tiled a N Then once we place this one piece here, we now have eight of the n minus one ways we can do. It's replace one piece. Okay. Now, you know, nothing we could do is to place a one by two piece horizontally and then another one underneath, Fix kind of trapped there. Okay, Well, then was left is a t by and minus two checkerboard which can be tiled aid. The n minus two ways. The third thing we could do is we could use our two by two tile here and again we get a two by and minus two checkerboard remaining which can be tiled am minus two ways. So if we combine this, let me get that eight of the end, which is the number of ways we can tile. A checkerboard can be expressed as a combination of the three kind of first move options. And so this is a the N minus one plus two A and minus two four and greater than equal to two. Okay. And so what we the initial conditions than for a equals 01 s o for a two by zero checkerboard, That could be tiled, uh, one way, which is no tiles. So a zero is one, and then a two by one checkerboard, which would just be a You think this could be tiled exactly one way, but putting the one by two piece written on it. So a one is also equal to one in case this is our our little formula here. Now we want to go ahead and find the roots characteristic equation. So if we let a n equal r squared and a and minus one equal o. R. And 80 and minus to equal one, if we put that into the expression we have here in red. It'll be r squared is equal to our my is too r plus two, um so we get our square is equal to our plus two. So we can that so tracked R plus two from both sides. So they are squared. My s r minus twos equals zero. You can factor. This is going to be equal to our minus two our plus one eso either ours equal to two or negative one. So now the other routes characteristic sorted out. We want to solve the recurrence relation. So remember, we have that a n is equal to a and minus one plus two and my eyes too. Ah, but a solution to the recurrence relation is going to be of the form A and is equal to a one are one Howard and plus a two are and are two to the heart of end. So this is gonna be Eagle Thio a one times to the end plus a two. It was negative one to be in. Okay, So what we know from our initial conditions is that one is equal to a zero cases. We put zero in here, then that must be equal to a one plus a two. We also know that one is equal to a one. So we put a one in here when we get to a one minus a two. So if we add these together, we'll get to on this style sort of ignore. This mill piece is equal to three times a one, which implies that a one is equal to 2/3. And then we can go back to one of these formulas say one is equal to a one plus a two put it in the value for a one. So one is equal to 2/3 plus eight two, which means that a two is equal to 1/3 so that we know those values we can put it in to the equation. Right? So we're looking at is in particular this equation since it's the last thing I was trying to fit up here, Uh, so then we get that A N is equal to 2/3 times two to the power, then plus 1/3 times negative one to the power of end s. So that is the solution to our recurrence relation

In this problem we have six plastic beads of different colors. So we have six plastic beads each of different color. And we have to find the number of ways in which blue and green beats never plays together. So let us first find the number of ways in which six beats can be arranged of different colors. Number of options that we have to fill the four spaces. Six because you're six different since one space is already filled by one of the colors. So for the second space we have five. This is four and so on until this as one. So the number of years. Mhm. To arrange. Uh huh. Six beads. Yeah. Is six factorial, which is 7 20. Now, let us find the ways in which blue and green beads are always together. Number of race such that blue green always together. So it means that blue green is one entity which is always together. And there are four other species that have to be failed. So this one has four options. Maybe from yellow, orange, black, and brown. This has three options. This has two options, and this is only one options. So the number of ways in which this can be arranged. This five factorial, keeping an account that busy can come in any place. Sophie factory. Now B. G itself can be interchanged as G. B. So we have to multiply it by two, which comes us five factorial into two is nothing but to 40 here, I've taken something like there are five entities with BG can be placed in five different locations. Others come like 4321 and so on. So this is five factorial into two which is 2, 14. So number of ways in which blue green never together will be equal to one minus two, which is 7 20 minus 2, 14, which comes as this is eight. And this is food. This comes us for it.

Here on this problem. We have six squares that we were going to be filling in with colors. What's what's that squares with spending all boxes here is we've got six whites here, we're going to fill it. We have 10 different colors that we can use here. But the catch is that we don't want consecutive squares to be filled with the same color Starting this out here. This means we have 10 options that we can have for that first square. You put any of those 10 there in that first one. Now, in the seconds where we only have nine options left. You just can't use whatever you put in that first square, Hearing that next where you still have nine options. It just can't be whatever was in that 2nd square. And the same for the fourth. You got nine options. It just can't be what was in that one right to its left. nine options there for the 5th. It just can't be what was right there to its left. And then lastly, nine options there in that last square. Yeah. And so to find the total number of options, we're just going to take all these and multiply them together. It's 10 times 1950 and 10 times 10 times nine of the fifth would give us 590,000, 490.

Okay, So palette of water core paints has three shades of green. Three shades of blue, two shades of bread, two shades of yellow in one shade of black. How many ways order to choose one shade of each one? What if three ways and choosing Bring three residues in blue. Two Ways of choosing Red. Two Ways of Chief Yellow on one way Coaches and Black. 23 times a very times two times two gives me 36 ways to choose one shade of each other.


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