Just to confirm on the last one, the elegance of the solution is that the empty dividers preserve the brand identity correct? in other words, you don't have to multiply by 4 because if you see in 1:09:23 that representation means only A's and that wouldn't be the same representation for only B's, C's, or D's because order of bars and lines matters here. Really impressed with the elegance of that solution.
Yes! For example, the representation of all D's would show the three dividers all the way to the left of the ten rectangles. All B's would have one divider all the way on the left, and two dividers all the way on the right. All C's would have two dividers all the way on the left, and one divider all the way on the right. Thank you! I can't take credit for coming up with this. I think the first time I saw a solution similar to this one was in a book by The Art of Problem Solving, but I'm not 100% certain.
Thank you, I was wondering whether to multiply by 4 or not by the symmetry of the problem, but now it makes total sense that the dividers preserve representation
Hey, Jeff, super elegant solution for the last problem. I'm curious to know how the factorial setup would change if we HAD to pick one brand (A, B, C or D) at least once or if it would at all. I would've offered a tentative setup here but I can't wrap my head around it. Thanks! (Tom)
Hi Tom! Let's say we HAD to choose at least one A-bar within our ten selections. How many combinations of our four candy bar varieties satisfy that condition? We can think of the problem this way: Get the assigned bar-A out of the way from the beginning. Then, we get to choose our remaining 9 candy bars from among four different options. Since the first candy bar is basically assigned to us, we can just remove it from the calculation. The problem boils down to counting the ways we could divided our 9 candy bar choices. There would now be 9 candy bars and three dividers, so 12 total objects. The number of combinations would be 12! / 9!3! Similarly, if we had to have at least one of each of A, B, C, and D, we'd really only have 6 free choices. So how many combinations would satisfy that restriction? Again, knock out those four assigned bars. Now we have to decide how to divide our 6 free choices. Six choices and three dividers means we have 9 total objects. The number of combinations would be 9! / 6!3!
In this problem there are two broad types of items: candies and dividers. We want to determine how the three dividers can sort the candies into 4 groupings; we will later call these groupings A, B, C, and D, but don't worry about that for now. Viewing the problem this way, we have a total of 13 items: 10 candies and 3 dividers. Now we need to figure out how many ways we can arrange those 13 items. This is where the 13!/((10!)(3!)) expression comes into play. We must account for the dividers because they will dictate how many of each candy bar are labeled A, B, C, or D. If you go to around 1:05:50, you can see me illustrate how these 13 items might be arranged. How we position the divider changes how many of each candy bar we will select: The number of As is determined by how many bars come to the left of the first divider. The number of Bs is determined by how many bars come between the first and second dividers. The number of Cs is determined by how many bars come between the second and third dividers. The number of Ds is determined by how many bars come to the right of the third divider. This can get a little overwhelming, so another way to confirm that the dividers play a role in our calculations is to try a scaled-down version of this problem: What if the vending machine only sold two types of candy bars, and you were going to buy a total of two candy bars. How many ways could you do this? The counting here is much simpler. There are only three options: AA, BB, AB. But notice that we can't arrive at a total of three options if we use only "2!". In this simpler version, consider that we actually have a total of three items: two candy bars and one divider. How many ways can you arrange those? 3!/((2!)(1!)) = 3 ways. If you feel comfortable with the simplified version, try this one: The vending machine sells three brands of candy bars. You are going to purchase four of them. How many combinations of candy bars could you purchase? If you can confidently arrive at "15," then the original problem should be within your grasp!
Correct me I am wrong but when order does not matter we use this formula C(n,r)=n!/(n−r)!r! But when order matters we use C(n,r)=n!/(n−r)! the video is reversing the two formulas
Hi Victoria, those two formulas look good to me. Could you let me know where in the video you believe the formulas have been reversed? I think there's a pretty good pair of examples on the screen at around the 28-minute mark that conforms to the formulas you've presented.
I think you meant to say P(n,r) = n!/(n-r)! P for permutation is where order does matter (e.g. password code), whereas C(n,r) = n!/(n-r)!r! Is where order doesn’t matter (e.g. arranging items in groups of a certain size). I don’t think the video reverses these two but it can be quite confusing as P and C are very similar with that tiny difference
Hi Joseph! The most concise way I can explain that the answer can't be (10!)/(4^10) is by pointing out that 10! is not divisible by 4^10. And since there must be an integer-number of ways to choose candy bars, this is going to be problematic. (Btw, 10! will be divisible by 4^4, but that's the limit. Might be worth confirming why this is true!) But I'm more interested in understanding where the 10! and the 4^10 come from in your expression. The 10! implies that at some starting point, we have 10 options to choose from, and then we have 9 options to choose from. And the number of options continues to decrease. I'm curious, when do we have exactly 10 things to choose from? And when do we have 9? Etc. It seems to me that, in a way, we always have 4 options/candy bar to choose from. soooo... That brings us to the 4^10 term, which would come up this way: if we were to make 10 decisions, and each decision had 4 options, we would get 4^10. For example, if we were to create a 10-letter code from the letters ABCD (repeats allowed), we would have 4^10 possible codes. But the problem with the candy bars doesn't care about the order in which we selected the candy bars--it only cares about how many of each candy bar we end up choosing with our 10 selections. Just grab 10 candy bars, put them in our shopping basket, and when you get to the cashier (or self checkout!) determine how many of each you took. 7As and 3Bs; or 4As, 5Bs, 1D; or 3As, 2Bs, 1C, and 4Ds; etc. This is equivalent to distinguishing between determining how many 10-character codes we could create with A's, B's, C's, and D's, and determining how many different assortments of A's, B's, C's, and D's could possibly end up in our 10-character code.
In the question "How many paths from X to Y", why are we considering the letters to be the same? Aren't they representative of the different squares, and thus unique? The first block is different from the second block. Getting confused here.
When moving from X to Y, we will move along 11 total edges of squares. Trace any path with your finger. 4 of the 11 moves must be in the downward direction (D), and 7 of the moves must be to the right (R). Suppose we take this path that kind of cuts through the middle of the grid: R-R-D-D-R-R-R-D-D-R-R. Does it matter if those first two Rs are swapped? Or the last two Rs? Or any two Rs? At each intersection on the grid, we only care about if our next move is a D or R, not which D or R from the batch we choose. All Rs do the same thing, and all Ds do the same thing. That's why we consider all the Rs to be the same and all the Ds to be the same. So while the first block is different from the second block, remember we aren't counting blocks; we are counting paths. Try setting up a 2x2 grid, and counting the number of paths from the top-left corner to the bottom-right corner. There are 4 total moves: 2 Ds and 2 Rs, and 6 total pathways. For this smaller set up, you can actually count all 6 paths relatively quickly, and 6 = 4!/(2! 2!)
Incredibly useful! Thank you!
Just to confirm on the last one, the elegance of the solution is that the empty dividers preserve the brand identity correct? in other words, you don't have to multiply by 4 because if you see in 1:09:23 that representation means only A's and that wouldn't be the same representation for only B's, C's, or D's because order of bars and lines matters here.
Really impressed with the elegance of that solution.
Yes! For example, the representation of all D's would show the three dividers all the way to the left of the ten rectangles. All B's would have one divider all the way on the left, and two dividers all the way on the right. All C's would have two dividers all the way on the left, and one divider all the way on the right.
Thank you! I can't take credit for coming up with this. I think the first time I saw a solution similar to this one was in a book by The Art of Problem Solving, but I'm not 100% certain.
Thank you, I was wondering whether to multiply by 4 or not by the symmetry of the problem, but now it makes total sense that the dividers preserve representation
Thanks Jeff... Long-standing doubt got clear. When to use Per or Comb. The thing is that "Order matters" :)
Hi Jeff - For the rearrange LIVER problem, do we subtract 1 to account for re-arrangement? I.e. considering only the values other than LIVER?
Aren’t these questions extremely rare in GMAT ?
Thanks Jeff. You are awesome!!
Hey, Jeff, super elegant solution for the last problem. I'm curious to know how the factorial setup would change if we HAD to pick one brand (A, B, C or D) at least once or if it would at all. I would've offered a tentative setup here but I can't wrap my head around it. Thanks! (Tom)
Hi Tom! Let's say we HAD to choose at least one A-bar within our ten selections. How many combinations of our four candy bar varieties satisfy that condition?
We can think of the problem this way: Get the assigned bar-A out of the way from the beginning. Then, we get to choose our remaining 9 candy bars from among four different options. Since the first candy bar is basically assigned to us, we can just remove it from the calculation. The problem boils down to counting the ways we could divided our 9 candy bar choices. There would now be 9 candy bars and three dividers, so 12 total objects. The number of combinations would be 12! / 9!3!
Similarly, if we had to have at least one of each of A, B, C, and D, we'd really only have 6 free choices. So how many combinations would satisfy that restriction? Again, knock out those four assigned bars. Now we have to decide how to divide our 6 free choices. Six choices and three dividers means we have 9 total objects. The number of combinations would be 9! / 6!3!
in the last question how does the divider come in place we should be dealing with factorials of candies
In this problem there are two broad types of items: candies and dividers. We want to determine how the three dividers can sort the candies into 4 groupings; we will later call these groupings A, B, C, and D, but don't worry about that for now.
Viewing the problem this way, we have a total of 13 items: 10 candies and 3 dividers. Now we need to figure out how many ways we can arrange those 13 items. This is where the 13!/((10!)(3!)) expression comes into play. We must account for the dividers because they will dictate how many of each candy bar are labeled A, B, C, or D.
If you go to around 1:05:50, you can see me illustrate how these 13 items might be arranged. How we position the divider changes how many of each candy bar we will select:
The number of As is determined by how many bars come to the left of the first divider.
The number of Bs is determined by how many bars come between the first and second dividers.
The number of Cs is determined by how many bars come between the second and third dividers.
The number of Ds is determined by how many bars come to the right of the third divider.
This can get a little overwhelming, so another way to confirm that the dividers play a role in our calculations is to try a scaled-down version of this problem:
What if the vending machine only sold two types of candy bars, and you were going to buy a total of two candy bars. How many ways could you do this?
The counting here is much simpler. There are only three options: AA, BB, AB. But notice that we can't arrive at a total of three options if we use only "2!".
In this simpler version, consider that we actually have a total of three items: two candy bars and one divider. How many ways can you arrange those? 3!/((2!)(1!)) = 3 ways.
If you feel comfortable with the simplified version, try this one:
The vending machine sells three brands of candy bars. You are going to purchase four of them. How many combinations of candy bars could you purchase? If you can confidently arrive at "15," then the original problem should be within your grasp!
Last question was very tough.
Correct me I am wrong but when order does not matter we use this formula C(n,r)=n!/(n−r)!r!
But when order matters we use C(n,r)=n!/(n−r)!
the video is reversing the two formulas
Hi Victoria, those two formulas look good to me.
Could you let me know where in the video you believe the formulas have been reversed? I think there's a pretty good pair of examples on the screen at around the 28-minute mark that conforms to the formulas you've presented.
I think you meant to say P(n,r) = n!/(n-r)! P for permutation is where order does matter (e.g. password code), whereas C(n,r) = n!/(n-r)!r! Is where order doesn’t matter (e.g. arranging items in groups of a certain size). I don’t think the video reverses these two but it can be quite confusing as P and C are very similar with that tiny difference
@@anishkrishnan9698 Good point, Anish!
In the last question, it doesn't say that she buys at least one of each of the 4 different brands, so how is it not (10!)/(4^10)?
Hi Joseph!
The most concise way I can explain that the answer can't be (10!)/(4^10) is by pointing out that 10! is not divisible by 4^10. And since there must be an integer-number of ways to choose candy bars, this is going to be problematic. (Btw, 10! will be divisible by 4^4, but that's the limit. Might be worth confirming why this is true!)
But I'm more interested in understanding where the 10! and the 4^10 come from in your expression. The 10! implies that at some starting point, we have 10 options to choose from, and then we have 9 options to choose from. And the number of options continues to decrease. I'm curious, when do we have exactly 10 things to choose from? And when do we have 9? Etc. It seems to me that, in a way, we always have 4 options/candy bar to choose from. soooo...
That brings us to the 4^10 term, which would come up this way: if we were to make 10 decisions, and each decision had 4 options, we would get 4^10. For example, if we were to create a 10-letter code from the letters ABCD (repeats allowed), we would have 4^10 possible codes. But the problem with the candy bars doesn't care about the order in which we selected the candy bars--it only cares about how many of each candy bar we end up choosing with our 10 selections. Just grab 10 candy bars, put them in our shopping basket, and when you get to the cashier (or self checkout!) determine how many of each you took. 7As and 3Bs; or 4As, 5Bs, 1D; or 3As, 2Bs, 1C, and 4Ds; etc. This is equivalent to distinguishing between determining how many 10-character codes we could create with A's, B's, C's, and D's, and determining how many different assortments of A's, B's, C's, and D's could possibly end up in our 10-character code.
In the question "How many paths from X to Y", why are we considering the letters to be the same? Aren't they representative of the different squares, and thus unique? The first block is different from the second block.
Getting confused here.
When moving from X to Y, we will move along 11 total edges of squares. Trace any path with your finger. 4 of the 11 moves must be in the downward direction (D), and 7 of the moves must be to the right (R).
Suppose we take this path that kind of cuts through the middle of the grid: R-R-D-D-R-R-R-D-D-R-R. Does it matter if those first two Rs are swapped? Or the last two Rs? Or any two Rs? At each intersection on the grid, we only care about if our next move is a D or R, not which D or R from the batch we choose. All Rs do the same thing, and all Ds do the same thing. That's why we consider all the Rs to be the same and all the Ds to be the same.
So while the first block is different from the second block, remember we aren't counting blocks; we are counting paths.
Try setting up a 2x2 grid, and counting the number of paths from the top-left corner to the bottom-right corner. There are 4 total moves: 2 Ds and 2 Rs, and 6 total pathways. For this smaller set up, you can actually count all 6 paths relatively quickly, and 6 = 4!/(2! 2!)
The neat riverbed inversely pass because chauffeur conversly squeeze till a second-hand certification. superficial, fallacious reason