Combinatorics/Permutations Just Isn't Clicking for Me

Question

I just can't wrap my head around it for some reason. I think I just need it broken down better than MGMAT book does, and also more in depth with practice problems and whatnot. Every time I look at a comination/permutation problem I get confused, bumble my through it, then usually get it wrong. Then I look at the answer it's like "duh". This is probably one of my weakest points in quant. I've always had trouble with stats/probability/combinatorics (and I'm an engineer!). Help!

mikemcgarry · Answer

Dear Devon , I'm happy to help. :-)

I did Physics in my undergrad, but when I first took probability and statistics in grad school, it was like Greek. I have a good grasp of it now, but it took some time --- it doesn't automatically follow from strength in general math. It's very much a thing of it's own. Here are a couple of blogs for starters: https://magoosh.com/gmat/2012/gmat-quant-how-to-count/ https://magoosh.com/gmat/2012/gmat-permu ... binations/ Remember that everything about both permutations and combinations can be derived from the Fundamental Counting Principle , discussed in that first link. Really, both the permutation formula and the combination formula are "shortcut" formulas that save a little time, so that we don't have to go back and rethink the whole situation from scratch, using the FCP. If you have questions about this, I would happily discuss it further. Alternately, it may help you simply to post one of the questions you got wrong, explaining how you thought about it originally vs. what you thought after the solution. Be careful reading those problem explanations ---- often the hardest part of a probability or counting problem is simply how the problem is framed: once it's framed properly, it's a piece of cake, but how do you know how to frame it? That's what we should discuss, but since that's different for each problem, we could only discuss that in the context of an individual problem. Does all this make sense? Mike :-)

Narenn · Answer

Agree with what Sir Mike said. Mastering the Combinatorics/probability concepts take some time. It is not that you can immediately master by memorizing certain formulas.

In addition to the articles provided by Mike, I would like you to consider the following articles also. Hope these will help you.

math-combinatorics-87345.html#p656598
permutations-and-combinations-simplified-150835.html#p1211661

Narenn

Bunuel · Answer

Theory on Combinations: math-combinatorics-87345.html DS questions on Combinations: search.php?search_id=tag&tag_id=31 PS questions on Combinations: search.php?search_id=tag&tag_id=52 Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html Theory on probability problems: math-probability-87244.html All DS probability problems to practice: search.php?search_id=tag&tag_id=33 All PS probability problems to practice: search.php?search_id=tag&tag_id=54 Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html Hope it helps.

Devon · Answer

Thanks for the response everyone. I guess I have trouble sometimes identifying how to go about solving a problem. For example, in MGMAT V5, p.50 they have the following problem:

A local card club will send 3 representatives to the national conference. If the local club has 8 members, how many different groups of representatives could the club send?

The book makes 8 slots, one for each representative, and then solves with filling the slots out with Y/N, etc.

My approach would be to make three slots, one for each representative sent, which would give 8*7*6=336 possibilities with repetitions. Now, in order to get to the correct answer I would need to divide by 3! (336/6=56). But I don't understand the reasoning behind dividing by 3!. I know that I need to divide to eliminate repetition, but where does the 3! come from in this instance? Can I even solve the problem in this manner?

Further, in their example the answer is 8!/3!5!. It says you need to divide by 3!5! to account for repeats, which in theory makes sense to me, but I don't understand why you're grouping those being sent and those not being sent as "identical" to give you the 3! and 5!. Is it just a rule of combinatorics, in a problem with less slots than options like this, to divide by factorials for BOTH groups (i.e. those going AND those not going)? How would this translate into how I tried to go about solving the problem?

Narenn · Answer

Well here you need to distinguish between Permutation and Combination. Consider the simple scenario. I want to give GMATCLUB TESTS free to any two of the three persons - You(de), your friend(f1), and your another friend(f2). So in how many ways can I give these tests?? It could be de-f1 OR de-f2 OR f1-f2 = Total Three ways. In mathematical term I would write it as 3C2 --------> 2 are to be chosen from 3 -------> \frac{3!}{(3-2)!*2!} ----------> \frac{3!}{1!2!} ---------> 3. WE CALL THIS AS COMBINATION Here I am selecting 2 persons from 3 persons. When you want to select r things from n things (Condition : r < n) you should use the formula \frac{n!}{(n-r)!*r!} Now Consider another scenario Now I want to make de, f1, and f2 seat on 2 chairs. I only have 2 chairs with me, So I will have to choose 2 persons from you. I will do that with 3C2. Now I have to decide which person to be seated on which chair. On first chair I can arrange any one of 2 persons and on another chair I can arrange the remaining 1 person. Total 2! ways Here I am performing two jobs in succession. Selection of 2 Persons from 3 Persons AND Arrangements of 2 persons at 2 places. -----------> 3C2 * 2! ------> 6 ways -----------> de-f1 OR f1-de OR de-f2 OR f2-de OR f1-f2 OR f2-f1 -------> WE CALL THIS AS PERMUTATION The direct formula to calculate permutations is nPr -------> n!/(n-r)! --------> 3!/(3-2)! -------> 3!/1! -----> 6 You will get to know these concepts very well, if you study the articles cited above thoroughly. Hope That Helps

mikemcgarry · Answer

Devon, You have to be very careful in thinking through the FCP in these situations. There are 8 members, say {A, B, C, D, E, F, G, H}, and we are going to pick three at random to send to the national conference. As you say, and the book says, we have eight slots --- 8 for the first slot, 7 for the second, 6 for the third, so 8*7*6 (BTW NEVER NEVER NEVER multiply a number like this out until all canceling it done! Leave it in un-multiplied product form! You are always making life harder than it needs to be if you wind up having to do math with a three-digit number such as 336.) Now, why isn't 8*7*6 the answer to the question? Because, among the three selected, order doesn't matter --- let's say that D & F & G were the three selected, well, the choices DFG DGF FDG FGD GDF GFD would count as six different "selections" in the 8*7*6 way of counting, but obviously, those are re-arrangements of the same three people. When we pick a group of n, we don't want to count rearrangements (i.e. permutations) of that group as different choices, so we have to divide by n! --- here, we divide by 3! = 6 The virtue of not multiplying out ---- (8*7*6)/6 = 8*7 = 56 ---- the only math we have to do is single digit math. If you do more than that on the GMAT Quant section, you are working too hard. Now, let's think about the expression \frac{8!}{(3!5!)} ----- think of this as \frac{8!}{(5!)}*\frac{1}{3!} ---- the first piece is the piece that cancels out the 5 people who won't be selected from the numerator --- the left fraction winds up as (8*7*6), and the right fraction cancels the permutations among the people who are chosen --- we wind up dividing by (3!), exactly as we had to do in the FCP approach, above. It's true there's a magic symmetric to the formula nCr = \frac{n!}{r!(n-r)!} , such that the number of combinations of 3 chosen from 8 has to be equal as the number of combination of 5 chosen from 8. This plays into the symmetry of Pascal's Triangle, about which you can read here: https://magoosh.com/gmat/2012/gmat-math- ... binations/ Does all this make sense? Mike :-)

OptimusPrepJanielle · Answer

First of all you need to have a basic understanding of the math behind these questions. I'm not talking about formulae, but how the formulae are created. The logic behind the math

Often, when you are stuck on a problem, it is not that the math is tricky, but that the context may be unfamiliar.The test writers are very aware of the vast number of possible contexts. And, while the math won’t change too much, they will often wrap familiar math concepts in misleading guises. Cracking the problem will be more a matter of choosing the correct approach than applying a given formula. Some such cases can involve beads on a garland, students in a class, triangles formed by connecting points etc. In order to get clarity on such questions, we have a targeted quant course for candidates preparing for the GMAT. You can find more details here: https://www.optimus-prep.com/gmat-quant-booster You can also opt for a well rounded course catering to all the needs of GMAT prep: https://www.optimus-prep.com/gmat-on-demand-course Feel free to take a 7 day trial of the courses we offer.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Combinatorics/Permutations Just Isn't Clicking for Me

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Combinatorics/Permutations Just Isn't Clicking for Me

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards