rehoot

5/16 sounds good. I haven't done these in a while, but I think the strictly mathematical way to do it is two rounds of binomial probability. The probability of getting exactly 3 heads would be 4C3 * .5^3 * .5^1 = 1/4 plus the probability of getting exactly one head is 4C0 * .5^1 * .5^3 = 1/16 so the total prob = 1/4 + 1/16 = 5/16.

I don't know what a jawan is, but I suspect that there were originally 8 of them in the problem--not 5. If there were only 5, then all 5 would need to be on the team.

This is a good answer for the puzzler in the Sunday paper, but I think that all GRE problems are restricted to plane geometry unless otherwise noted. Can I make this assumption that we are stuck in a plane when the problem gives us figures of plane geometry???? If we can't make this assumption, then some of the other geometry problems are called into question.

Without getting too focussed on the story behind the problem, we have some criteria that has to be met in the $250 problem: 1) all 250 of the $1 bills must be distributed to the seven people 2) every person must get at least $1 (this doesn't change the anser here ,but it affected the original problem at the top of this thread) 3) some people can get more than others, which is equal to saying that not everyone needs to get the same amount of money Now we can look at options A and B to see if they introduce new criteria. When you first read the problem, it sounds like option B allows more ways to distribute the money since you don't have that extra restriction about the $34 minimum, BUT when we go back to the first criteria (all 250 bills must be distributed), we find that there is no way to distribute all of the money without paying at least one person $34 or more. Try to show me one case where you can meet the 3 criteria and not pay anyone $34 or more (maybe I made a typo??) I can't find one. The way I tried to find a scenario that fits option B that isn't included in option A was to distribute as much of the money as I could without hitting that $34 limit. That is when I found that $33 to each person yields only $231, and to meet the criteria I have to give away the full $250, so one or more people will have at least $34 because we have an additional $19 (250-231) to give away and everyone already has $33.

Also, priyankagre2005 posted a link a couple weeks ago that explains the above: http://www.www.urch.com/forums/showpost.php?p=206067&postcount=19

Yea, I'm wondering if these were from the old analytical section. Everyone knows that there is no longer a GRE analytical section (on the general test), right?

I also tried breaking the problem down to its most fundamental representation by breaking it down to two people, but I used names: {Bill, Hillary} {Monica, Joe} I got stuck because if Monica shakes Bill's hand I wasn't sure if he was really shaking her hand or if his hand was merely being shaken by her [clap]

I saw this a couple weeks ago where the money was distributed across two people, and at first I thought the problem here would be really hard until I saw bondsergey's note that in any scenario, there has to be one person who will get $3 or more. If you try to find a scenario where nobody gets $3 or more, you might try giving the biggest number below 3, which is 2,and you get 2+2+2 which is illegal since the total has to be seven, so it becomes 2+2+3. The result is that no matter what you get for the true number of combinations for option A (at least one person gets $3 or more), it will be exactly the same as the answer for column B since both must contain at leas one person who gets $3 or more. I can now solve the following problem in a matter of seconds (at least I think I can and I hope I don't make a mistake and thereby make a complete fool of myself ?:p : Q: Joe has 250 one-dollar bills to distribute to 7 people for payment for work done, and he must distribute all of the money. Each person must get at least $1, but if some people worked extra hard, they can get more than the other people. Which is greater: A) the number of ways to distribute the money where at least one person gets $34 B) the total number of ways to distribute the money ANSWER: Lets try to find a scenario where at least one person does not get $34 to see if we have extra scenarios to research. I'll try giving everybody one dollar less than $34, which is $33. When I do this I find that 33 * 7 = 231, which is not enough to distribute all of the money, so at least one person must get paid $34 or more. This means that there are no additional scenarios for option B that are not already counted under option A. Without even counting the number of permutations, I choose C, they are equal. Did I make a fool of myself, or is the above true

This is my attempt to contribute to the sensless wasting of data storage.

Yes, I think I dropped the extra "5!" above, so 216 looks good. thx for checking

I'm not sure on the wording of the problem, are you saying for number 1 that there will be only two books on the shelf with one being a physics book and one being a math book? If that is the case that I would agree with the 2 * 5c1 * 4c1 answer (= 40). The "2 *" part comes from the arrangement of: 1) physics book first (leftmost) on the shelf and 2) math book first (leftmost) on the shelf. For question 2, I agree with MWM also.

Q. How many five digit numbers can be formed using the digits 0,1,2,3,4 and 5 which are divisible by 3, without repeating the digits? A number that is divisible by three has the sum of its digits divisible by 3. The problem says that you can't repeat the digits. I think we could identify the 5-digit sets that have the sum of the digits divisible by 3, then find the permutations on each unique set. My attempt to find the sets is this: {1,2,3,4,5}, {0,1,2,4,5} so I get 2*(5p5) - 4p4 = 120-24=96 where I think that 4p4 might represent the number of permutations that have a leading zero, and we can't count those. Q. From 4 officers and 5 jawans, in how many ways can 6 be chosen to include exactly one officer. I don't know where the 8c5 came from. It would seem to me that the answer is 4c1 * 5c5 If you are choosing 6 people and exactly one of them is an officer, then you are always choosing all 5 of the jawans. I'm assuming that a jawan is a deputy or other such employee.

The general idea is to identify a pattern then count how the value is repeated. I built a table with two columns: (pardon my dots, but the html changed my multiple spaces to a single space) range.........nbr of ones -------.........------------ 0-9............1 10-19.........11 20-99.........8 100-109......11 110-119......21 120-199......88 200-299......20 (same as 0-99) You could reduce it a few steps by seeing that 0-99 had 20 ones, 100-199 will have 100 more than that, and 200-299 will be the same as 0-99 giving 20+120+20=160

For the mwmwmwmw problem, it is the women who must be separated. To separate the women, you need at least n-1 men, but you may have many more men, for example, you could have 5 men and 3 women. If you create a template for where the women might go, it may look like the following where the underlines are "potential" spots where a woman could be seated: _M_M_M_M_M_ So there are (M+1) potential seats for women that will be populated with N women. The number of combinations that tell you which parts of the template will be filled with women is (m+1)Cn (chosing n "seat index numbers" from a list of (m+1) values) which is: (m+1)! / (m+1 - n)!n! ...because the generic equation for combinations is n!/(n-r)!*r! (where the n here is not related to the current problem). We still have to count the number of permutations for seating the men in the men's seats and the women in the n number of womens' seats independent of where those seats are -- we will always have m seats for men and n seats for women regardless of where they are in relation to eachother. For the men, the number of permutations is m! and for the women it is n!. So the full equation would be the total number of men permutations times the total number of women permutations times the number of combinations related to where we can please the women's seats in the template above. This is now: m!*n! * (m+1)!/(m+1-n)!*n! and the n! values cancel, and after I rearange the letter in the denominator to standard form, I get: m! * (m+1)!/(m-n+1)! That seems to explain it (?)

these numbers look too ugly for the GRE, but I get... the sides are in the ratio 2:sqrt(5):3 via Pythagorean The real sides are in that ratio times a scaling factor that I'll call "x". The area is bh/2 = 2x * sqrt(5)x/2 which was given to be 23. Simplify that one step to get: x^2 * sqrt(5) = 23, keep simplifying... x^2 = 23/sqrt(5), now take the square root of each side: x = sqrt(23/sqrt(5)) (so the 5 is effectively to the 4th root) Remember that x is the scaling factor. The short side was 2x which is 2* sqrt(23/sqrt(5)) (you can simplify that if you are bored). As a check, I'll find the height: sqrt(5)x which is sqrt(5) * sqrt(23/sqrt(5)) To verify the area, try (1/2)bh = (1/2) 2x * sqrt(5)x and get: sqrt(23/sqrt(5))^2 * sqrt(5) =23/sqrt(5) * sqrt(5) so the sqrt(5)'s cancel, leaving 23 as a check.

I would guess "E" since the others don't seem to be divisible by 24. I'm just looking at the progression of the "4" in 24 -- I'm looking at the pattern of the multiples of 4. ... I just did fiddled with the calculator and found 65520 is divisible by 24.

OK, 163 it is. The problems has now been fully illuminated. I didn't understand the notation of the earlier posts.

I am not confident that I have the answer to the TENNESSEE problem, but I did some research on it and here is what I have... To start with an easier example of permutations that have a duplicate letter, look at the "moon" example here: http://home.blarg.net/~math/permutations2.html when you take a subset of the set that contains repeated letters, see the "zoo" example here: http://math.youngzones.org/P&C.html When you have lots of letters and take r letters from a set of n where many letters are replicated, you have to break the problem into pieces (look for the BANANA example): http://www.mansw.nsw.edu.au/members/reflections/vol21no1_larkin.htm My attempt to break the problem into pieces is as follows (I haven't done this before, so I could be way off): A) 4 unique letters with no duplicates: 4!=24 permutations B) 4 letters, two of which are duplicates (like TENSS, TENN, TSEE) T, E, {S, S} T, E, {N, N} T, S, {E, E} T, S, {N, N} T, N, {S, S} T, N, {E, E} With each group having 4!/2!=12 unique permutations (see the links above for the equation to remove the duplicates from the permutation calcualtion) C) 4 letter words with two sets of duplicated letters: {S, S}, {N,N} {S, S}, {E, E} {N, N}, {E, E} with each set having 4!/(2!*2!)= 6 unique permutations D) 4 letters, three of which are duplicates (only E has enough for this): T, {E, E, E,} S, {E, E, E,} N, {E, E, E,} With each set having 4!/3!=4 unique permutations Total of A, B, and C is 24 + 6*12 + 3*6 + 3*4 = 126 ? Remember that this is my guess. As for the number of COMBINATIONS of four-letter words, remember that the order of the letter is irrelevant. There is one combination that has 4 unique letters, then one combination for each of the sets above = 13 combinations total ?

If that were true, which it isn't, then would the probability of both hitting be: 2/3 + 4/7 = 26/21 = 123.8%? No. The chance that they both miss is 1/3 * 3/7 = 3/21

Good question. I found a page that seems to explain it clearly. Start at the top and then read the example for "Moon" http://home.blarg.net/~math/permutations2.html My explanation for the "PreUniversity" question is that there are two copies of letters "e," "i,", and "r", and since the double copies of the lettes are identical, you can't count them twice (read the "moon" example in the link). So you have to divide the 13! by (2!*2!*2!). I dropped the hyphen from the spelling since it was confusing me and not included in your 13!. I don't see any duplicate lettes in "simpleton," so my answer for Q1 above is 9p9 (which also = 9! and means that you have to use all the letters in the word. My answer for Q2 is less certain since I'm not exactly sure if they meant to take all 9-letter groupings (9p9), then all the 8-letter groupings (9p8) and so on. If you can take all the 9-letter permutations, plus all the 8-letter permutations and so on, then I'll stick with my original answer.

Deren, regardless of the intent of the original question, you have a good point that I would have overlooked: Deren said: "I don't agree with u ... the question does not tell u that these numbers are consecutive... => a=-10 b=-9 c=5 d=9 e=10 so; median=5 and mean would be = [-10+(-9)+5+9+10 ]\ 5 = 1 this means; Median > Mean"

Kronique, When I first read that problem I thought that "72.42" was a shortcut for "72* 42" because other people have used that notation, but it really is 72.42. Then, you group then equations in a factored mode so that you can how they parallel eachother: I'll do it in pieces LHS (left-hand-side): 72.42 = 72 + 42/100 RHS: k(24 + n/100) = 24k + nk/100 Now factor a 3 out of the LHS: LHS: (24*3 + (14*3)/100) Now you can see how similar they are: 24k + nk/100 = (24*3 + (14*3)/100) so you can equate the terms: 24k = 24*3 and nk/100 = (14*3)/100 From the first grouping, we see that k=3, then put that into the second grouping and we see that n=14

1. I think this may be a case of charientism, but nevertheless, If you take all the letters of a word, there are 9p9 permutations: n= the number of things in the set (9) r= how many things we are putting in each group (9) p=n!/(n-r) 9!/(9!-9!) = 9!/0! = 9/1 = 9! = 362,880 (yes, 0! = 1) 2) I'm not positive, but this question may be asking for all the permutations, meaning 9p9 + 9p8 + 9p7+ 9p6+ 9p5 + 9p4 +9p3 + 9p2 + 9p1 You can do the math. I don't think this would be a GRE question since I don't think you would have to do that many calculations.

Yes, Kronique is correct, I should have said (1/18,009,460) By the way, if the chance of winning the jackpot was 1/8, I would win at almost every week and I wouldn't be wasting my time on the GRE!

Look at it this way. If both have to miss, and the probability that the first person will miss is only 1/3, then already the chances of both missing are well under 50% (and the proposed answer of 13/21 is more than 50%). After the 1/3 chance that the first person will miss, the next dart-thrower usually hist the target, so the chance of missing is only (1-4/7) = 3/7. The new cumulative chance of missing is 1/3 * 3/7 which is 3/21. This equal about 14% chance that both will miss. Using percents, the first thrower will miss only 33% of the time, the next person will miss only about 43% of the time, and this will reduct the 33% number to .33 * .43 = .14, or about 14%.