GRE Quant Ep 18: Combinations & Permutations

[@ramachandrarao6469]

Very helpful. Great work. I could remember all the stuff I have studied in my engineering entrance exam about P and C again easily after going through this lecture. Thanks a lot .

[@GRENinjaTutoring]

I'm glad that it was helpful, and brought back fond memories of engineering exams. :) Have fun studying, and please feel free to keep us posted on your progress!

[@saracyber1]

Thank you so much, Ninja! Your channel helps me a lot!

[@RishabhJain-cy4ss]

really helpful video! thank you so much

[@GRENinjaTutoring]

Thank you so much for watching!

[@NYJanus]

Super helpful

[@GRENinjaTutoring]

Thank you so much!

[@HadushHailu]

Your videos are awesome, Ninja! I have one question. In the question number 1, why would order matter for creating a menu? after all whatever order it comes, it is still the same menu?

[@GRENinjaTutoring]

Thank you for the kind words! In this question, we're told that Saheera "will eat the ice creams in the order she chooses them." This means that choosing vanilla, banana, strawberry, and chocolate in that order would create a different menu for Saheera than choosing chocolate, strawberry, banana, and vanilla in that order. If this was just a question about choosing four ice creams from the eight available, then the order wouldn't matter. However, the wording of the question tells us to take the order into account in this question. I hope that helps!

[@reynardonainggolan1298]

awesome

[@GRENinjaTutoring]

Thank you again! I'm glad that you're enjoying these so much. Have fun studying!

[@martynadydyk9714]

I did Q6 correct, I don't know how, I have no clue what I did, but I got the answer right.

[@ap1038]

Awesome video! Just one question, how can you justify no repeated digits on question 4? As you put 8 choices on second spot and 5 on third spot for 800s? And 8 digit choices on second spot and 4 on third spot for 900s?

[@GRENinjaTutoring]

I'm sorry, I don't think I really understand your question. Question 4 told us to find the number of odd, three-digit integers that have no repeated digits. It wasn't a case of justifying it, that's just what the question asked us to do. I hope that helps, but I suspect I probably haven't answered your question here. If I haven't answered it, could you please rephrase the question, and I'll do what I can to help. Thank you!

[@ranijirjees8589]

@@GRENinjaTutoringI was confused on this too. If you used the 8 and all the odd numbers wouldn’t you only have 0,2,4,6 to use left for the middle for it to not be a repeated digit or am I thinking of this wrong.

[@GRENinjaTutoring]

@@ranijirjees8589 Just as an example, let's assume we used 8 for the first and 1 for the final digit. If we did this, we could choose any one of 0, 2, 3, 4, 5, 6, 7, or 9 for the middle digit. The number would still be odd because the final digit is 1, and we wouldn't have repeated any digits. This means there are 8 options for the middle digit. We could repeat this process and swap the odd number we used in the final digit to give us the other options. This means there is 1 option for the first digit (it has to be 8), 5 options for the final digit (it has to be an odd number), and 8 options for the middle digit (what's left). I hope that helps!

[@joeboxter3635]

In the at least one enginer, why does 3C1 * 6C3 not work? You have to have 1 engineer. And there are 3 choices for engineer. Now you have 6 people and 3 openings in team which is 6C3. So why does that not work?

[@pezpeculiar9557]

17:19 Why does 6C3 not work? I thought effectively you're building a team of 3 people if you assume one of the four is already an engineer

[@GRENinjaTutoring]

To use 6C3 as the method to find the number of teams, we've looked at the 3 engineers, put one of them in the team, and put the other two in a group with the 4 sales personnel so we can do the 6C3 operation. By doing things this way, we've effectively said that the first engineer we chose and put in the team before doing 6C3 with the others MUST be in the team. We build the rest of the team around them and they MUST be in the team. However, there can be teams that do not involve this first engineer that we'd miss if we only did 6C3 to find the solution. To get to the right answer, we also need to find the number of teams that include at least one engineer but do not involve this first engineer. We could repeat the process by excluding the first engineer, choosing a second engineer that MUST be in the team, putting the final engineer with the sales personnel, and doing 5C3 to find the other members of the team. But there's still more to do: we need to repeat this process one more time to find the number of teams that include at least one engineer but do not include the first two. To do this, we'd say the final engineer MUST be in the team, exclude the first two, and choose 3 sales personnel from the available 4 to make up the rest of the team. There are 4C3 ways of doing this. This means we'd have a total of 6C3 + 5C3 + 4C3 = 20 + 10 + 4 = 34 possible teams, giving the same answer that Harry found in the video. I hope that helps!

[@AdedoyinAdekunle-z1k]

Wow, this is huge. Thank you so much for the sacrifices and efforts put together to make GRE prep easy. I actually need some clarity on question 5. I tried to use the method in question 3 where the result of 4C4 was subtracted from 7C4 to get 34. But in question 5, it says "at least 2 must have a long wheel", I tried subtracting the results of 4C4 from 9C4 but got stuck in getting 105 as the answer. Could you please help me on this?

[@GRENinjaTutoring]

Thank you for your kind words! I'm so pleased these videos are helping you. The difference between Q3 and Q5 is that Q3 asks us to find the number of teams that include "at least one" engineer, while Q5 asks us to find the number of ways we could choose the vans with "at least two" of the vans having long wheelbases. We have to approach these two questions differently because of the difference in phrasing. In Q3, we could find the total number of possible teams and then subtract the number of teams that include no engineers. Sadly, Q5 is not quite so simple. In this question, we could find the total number of ways of choosing four vans from the nine available, but then we'd have to subtract the number of ways of choosing four vans with none of them having long wheelbases AND the number of ways of choosing four vans with one of them having a long wheelbase. If you did this, you'd get 126 - 1 - 20 which would give us the 105 that's the answer to this question. Alternatively, as Harry showed in the video, we could find the number of ways of choosing the vans with two, three, and four of them having long wheelbases, and then sum these results to find the final answer. I hope that helps!

[@julianamartinez9737]

@@GRENinjaTutoring How would the formulas work for Q5 if finding the total number of ways then subtracting the others? Im trying 4C1 for "choosing 4 vans with one of them having a long wheelbase" but that's giving me 6 instead of 20 like you said. 4C0 would be for the "choosing 4 vans where none have long wheelbases"?

[@GRENinjaTutoring]

@@julianamartinez9737 Let's set up the situation: we have 9 vans, 5 of which have a long wheelbase and 4 of which have a short wheelbase, and we want to find the number of ways of choosing 4 vans such that 1 of them has a long wheelbase. To do that, we can do 5C1 to find the number of ways of choosing one van with a long wheelbase and 4C3 to find the number of ways of choosing three vans with a short wheelbase. This will give us 5 and 4, and we multiply these numbers together to give us 20. For an explanation of why we multiply the 5 and 4 together, check out the explanation for Q2 in the video as there's an identical scenario in that question. I hope that helps!

[@julianamartinez9737]

@@GRENinjaTutoringthank you!! it’s still a little confusing but i’ll do some more practice problems!

[@rheak.5470]

Hi, for q3, the method presented is definitely more efficient but just wondering if (4C3 x 3C1) + (4C2 x 3C2) + (4C1 x 3C3) = 34 is also theoretically, correct because that was the first one that came to mind when solving. Thanks!

[@GRENinjaTutoring]

Yes! You've followed the method Harry described in blue at the start of the explanation. It's a perfectly valid way of doing this question and if you're comfortable with finding the values of the choose functions, it won't take too long. It's just not quite as efficient as the method shown in the explanation. If your method was the first one that came to mind and you're comfortable with solving it that way, you'll do just fine in this question. Thank you for adding an alternative solution path! I hope that helps!

[@rheak.5470]

@ alright thx!

[@Kings_707]

why is it 1680 but not 70?

[@GRENinjaTutoring]

The important thing in Q1 is to notice that the order in which Saheera chooses the ice cream matters. If she chose vanilla, then strawberry, then chocolate, then banana she would have a different menu than if she chose banana, then chocolate, then strawberry, then vanilla. When the order matters, it's best to use the slot method demonstrated in the video. If we do that, we'll find the answer to this question is 1680. If the order in which Saheera chose the ice cream did not matter, then we could use the choose formula. If we did 8C4, we'd get an answer of 70. I hope that helps!