For the 3rd quiz question, is it semi-intuitive or just luck to know that (even * even = even), (even*odd = even), and (odd*odd = odd) therefore there is a 2/3 probability that you will get an even number when multiplying evens and odds.
Hi for question 3, I approached it differently than your approach. I solved the total possibility to make groups of 4 from a pool of 7 which I got 35 ways (7!/(3!*4!)). I then subtracted from 35 the number of ways to have Ben and Ann in a group (out of 4 slots, first two are taken by ben and ann, so remaining two 5*4 = 20) thus I have 35-20 = 15 ways to form a group without Ben and Ann, which is different from your 5. Please advise. Thanks !
I did some thought and came to the conclusion that perhaps my error was when selecting the 2 more contestants after ben and ann, but even when I take it as a combinatorics approach the groups that contain both Benn and Ann (minus the overlaps from difference in order of picking) comes down to 10 (5!/(2!)(3!)). So the answer would still be different than 5 as 35-10 is 25. I cannot figure out why my approach would lead to a different answer than your approach logically nor can figure out a math flaw...
Hi Charles, can you please give some more examples of the harder questions? Or link me to a video where I can further practice those? i.e. If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
Hi Charles, is there any way I can practice the harder type questions? Need to get a really good score for my program. i.e. If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
With a risk of sounding like a self-promotion, the question pool of GMAT Club tests (especially quant) is generally considered hardest. You may want to test it out by taking a free test.
Now that GMAT Focus has cut the Questions down to 21 on PS, there is basically only going to be 1 (at most) probability question. There is a possibility to face Probability and combinations on the DI. You need to have a general understanding of the concept and how to solve fairly simple questions. Usually it is not worth spending weeks in Probability... it gets complex and I only get lost 😂 -BB
On question 1 - How come when working out the number of groups of 3, you don’t do 10x9x8x7x6x5x4x3x2x1 divided by 3x2x1? I saw you only did 10x9x8 and just trying to figure out why
For the 3rd quiz question, is it semi-intuitive or just luck to know that (even * even = even), (even*odd = even), and (odd*odd = odd) therefore there is a 2/3 probability that you will get an even number when multiplying evens and odds.
Hi for question 3, I approached it differently than your approach. I solved the total possibility to make groups of 4 from a pool of 7 which I got 35 ways (7!/(3!*4!)). I then subtracted from 35 the number of ways to have Ben and Ann in a group (out of 4 slots, first two are taken by ben and ann, so remaining two 5*4 = 20) thus I have 35-20 = 15 ways to form a group without Ben and Ann, which is different from your 5. Please advise. Thanks !
I did some thought and came to the conclusion that perhaps my error was when selecting the 2 more contestants after ben and ann, but even when I take it as a combinatorics approach the groups that contain both Benn and Ann (minus the overlaps from difference in order of picking) comes down to 10 (5!/(2!)(3!)). So the answer would still be different than 5 as 35-10 is 25. I cannot figure out why my approach would lead to a different answer than your approach logically nor can figure out a math flaw...
Hi Charles, can you please give some more examples of the harder questions? Or link me to a video where I can further practice those?
i.e. If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
Hi Charles, is there any way I can practice the harder type questions? Need to get a really good score for my program.
i.e. If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
With a risk of sounding like a self-promotion, the question pool of GMAT Club tests (especially quant) is generally considered hardest. You may want to test it out by taking a free test.
in general how many questions based on Probability and Combinations appear in GMAT examination?
Now that GMAT Focus has cut the Questions down to 21 on PS, there is basically only going to be 1 (at most) probability question. There is a possibility to face Probability and combinations on the DI. You need to have a general understanding of the concept and how to solve fairly simple questions. Usually it is not worth spending weeks in Probability... it gets complex and I only get lost 😂
-BB
On question 1 - How come when working out the number of groups of 3, you don’t do 10x9x8x7x6x5x4x3x2x1 divided by 3x2x1? I saw you only did 10x9x8 and just trying to figure out why
10!/3!(10-3)! is formula and (10-3)!= 7!= 7x6x5x4x3x2x1 which is cancel that's why only 3! is remains