![]() ![]() ![]() It’s much easier to see that there are 299 numbers in the latter case.) (Another way to see this is that the range between 701 and 999 is the same as the range between 001 and 299, since we simply subtracted 700 from each number, keeping the range identical. So there are (999-701)+1 = 299 numbers in total without restrictions. To get the total number of terms, we must subtract the two numbers then add one to account for the end point. Since it says “greater than 700”, we will not include 700.) ![]() In this case, that’s just all the numbers from 701 up to 999. If we can figure out the total number of permutations without restrictions and subtract out the number of permutations in the two situations just listed, we will have our answer.įirst, let’s get the total number of permutations without restrictions. the ones not desired )? Well, if you take a little time to think about it, there are only two other possibilities: The situation desired is 2 digits equal and 1 different. This will be very akin to how we handle some GMAT probability questions. But as with the previous problem, what if we examine conceptually what doesn’t work? You could try writing up examples that fit the description, such as 717, 882, 939, or 772, trying to find some kind of pattern based on what does work. This is a classic example of a problem that will tie you up in knots if you try to brute force it. Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ? As always, give it a shot before reading on: This is why tougher combinatorics questions are more likely to involve restrictions. Īs you can see, this is not about a formula or rote memorization but instead about logic and analytical skills. Subtract that from the original 60, and we have: 60 – 24 = 36. That produces a total of 6*4 = 24 permutations in which the two I’s appear side by side. Situationally, how would I outline every possible way the two I’s could be adjacent? Well, if I imagine the two I’s grouped together as one unit, there are four possible ways for this to happen:įor each one of these four situations, however, the three remaining letters can be arranged in 3*2*1 = 6 ways. This is where things become less about math and more about logic and conceptual understanding. However, we now want to subtract out the permutations that involve the two I’s side by side, since this condition is prohibited by the problem. ![]() Using the principle discussed in our Permutations with Restrictions post, this would produce 5! / 2! = 60 permutations. In the word “DIGIT,” there are five letters and two I’s. In this case, we’ll pretend this problem has no restrictions. what is not allowed) and then subtract that number from the total number of permutations without any restrictions, wouldn’t we then be left with the number of ways in which the two I’s would not appear together (i.e. It seems counterintuitive at first, but if we consider the number of ways in which the two I’s can appear together (i.e. So how do we handle this? Well, in many cases, it’s helpful to set aside what we want and instead consider what we don’t want. Put differently, they are not allowed to be adjacent. However, we’re suddenly told that the two I’s must be separated by at least one other letter. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?ĭid you catch the restriction? Up until the end, this is a standard permutation with repeats combinatorics problem, since there are five letters and two repeats of the letter ‘I’. The letters D, G, I, I, and T can be used to form 5-letter strings as DIGIT or DGIIT. To illustrate this directly, let’s take a look at the following Official Guide problem: Now, we’ll see what happens when permutation problems involve conceptual restrictions that can obscure how to approach the math. So far, we’ve covered the basics of GMAT combinatorics, the difference between permutations and combinations, some basic permutation and combination math, and permutations with repeat elements. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |