bmwhype

it depends on your program. u either have to take GRE or GMAT.

start with the first term, which is the first integer, 1 n/n+1 = 1/2 1/2 * 2/3 * 3/4 * etc... 1/2 all the way to 10/11 cancel out the terms by simplifying and we get 1/11 D

very nice. OA is 48

the question asks for E and J to sit next to each other. this is why we consider EJ and JE. if it asked for E to sit in front of J, it would be as u described it.

its a clear E. Use huge numbers, 100 and -100 and it will be clear.

krusta is right. that is the p(at least one)

simple question. 3-2+1= 2 2 elements to arrange arragnement within the group = 2! 2!2!=4

just google the probability and combinations. i find the attachments subpar and some of the answers are a bit faulty.

thanks krusta for your input. i really appreciate your efforts (so that i know im not wasting my time posting questions) these are the answers. made up the q myself. 84) There are 6 different colored balls along with 6 corresponding colored boxes. a) What is the probability that exactly 1 ball will go into its corresponding box? 6 * 1/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1/1 = 1/5 Explanation 6 balls --> call them Ball 1, Ball 2 …etc 6 boxes Assume that Ball 1 goes into its CORRECT corresponding box Ball 1 has 1 possible CORRECT box out of 6 boxes.--> 1/6 Now every subsequent ball has to be WRONG. Ball 2 has 1 CORRECT box of 5 boxes. Ball 2 has 4 WRONG boxes out of 5 boxes --> 4/5 Ball 3 goes in wrong box--> 3/4 Ball 4 goes in wrong box--> 2/3 Ball 5 goes in wrong box-->1/2 Ball 6 goes in wrong box-->1/1 Joint probability = 1/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1/1 = 1/5 This joint probability is where only Ball 1 is correctly placed. The stem did not say which ball is correctly placed. Therefore, we must multiply the joint probability by 6 to get all 6 possible scenarios. Therefore, 6 * 1/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1/1 b) What is the probability that exactly 2 balls will go into its corresponding box? 6 * 1/6 * 1/5 * 3/4 * 2/3 * 1/2 * 1/1 = 1/20 c) What is the probability that exactly 5 balls will go into its corresponding box? 6 * 1/6 * 1/5 * 1/4 * 1/3 * 1/2 * 1/1 = 1/120 d) What is the probability that all balls will go into its corresponding box? Ball 1 goes into box 1, ball 2 goes into box 2…etc. This scenario let’s all the balls go in their respective boxes. We don’t multiply by 6 because each ball goes exactly where its supposed to. Whether ball 1 goes in the box 1 first or ball 2 goes into box 2 first, the final outcome is the same. 1/6 * 1/5 * 1/4 * 1/3 * 1/2 * 1/1 = 1/720 1/6! = 1/720 e) What is the probability that no balls will go into its corresponding box? 1 – all = none 1 – 1/6! = 719/720

6 guys (let's call them ABCDEF) sit around a circular table. If the AB must always sit together, and C must face the door, how many arrangements are possible? please explain your work...

There are 6 different colored balls along with their 6 corresponding colored boxes. What is the probability that none are correctly placed in their correct respective boxes?

can someone show the work?

thanks. the wording is quite tricky, but it basically implies RBY is different from RBG, which is a property true of any combination problem.

In Finn's preschool class, each student is assigned a unique color palette of 3 colors for a finger-painting project. Different students may have 1 or 2 colors in common, but no 2 students have the same 3 colors. If there are 10 students in the class, how many different colors are required? (A) 4 (B) 5 © 6 (D) 7 (E) 8 Please explain your approach (preferably not backsolving) and logic

thats correct. figured it out yesterday. the question asks how many MUST, implying the least number of people, have all 4.

What is N in each statement? 1. (n+2)! is divisible by 36 2. (n+2)! is divisible by 49 in factorial divisibility questions, do we look for the highest factors or the highest prime factors? in #1, do we look for the lowest number of where we can derive 5 instances of 2? 36 = 6*6 or 36 = 2^5 in #2, do we look for numbers with one instance of 7, or two? (N+2)! = 7 or (N+2)! = 49

how many draws can we make from the word POOL?

sorry but that was not my intent

can someone explain the logic? how would u acccount for ppl who has 2 of a kind, 3 of a kind, 1 of a kind and finally all 4?

the OA was already given as 12. he doesnt understand the concept

yes they are the same. they are pretty good for sharpening some skills, specifically number prop, divisibility, rates and combos.

GMATCLub Challenges some of the rate questions are ridiculously hard and wont ever be tested...

clearly C. median depends on the number of elements in the set. there are 5 elements here so n or k must be 10, place them in increasnig order and you'll see that n is the median because it is bigger than k, n =10

for some reason, i can paste the excel table onto the post field but it converts to something else.. M~MS.25(156)70~s156248

i think double set matrix is much faster for me. i dont have to worry about overlaps because they are very clear cut. 31/248 1/8