krusta80

If the order of the individual seat locations matters, then you are correct: the answer is 12! Note that this number can also be derived as follows: # of Groups of 7 (and 5 left over) that can be created from a population of 12 = 12C7 = 12C5 = 12! / (7! * 5!) # of unique arrangements of 7 people in 7 distinct chairs = 7! # of unique arrangements of 5 people in 5 distinct chairs = 5! # of total unique seating arrangements = 12C5 * 7! * 5! = 12!

Let A denote the number of Albs Let B denote the number of Berks Let t denote the number of ticks Let c denote the number of clicks From the given portion of the question, we know the following: t = floor(A/3) c = floor(B/4) total = t*c = 77 Since t and c must be non-negative integers whose product is 77, there are four possibilities: t c 1 77 --> albs can be 3,4, or 5 / berks can be 308,309,310, or 311 77 1 --> albs can be 231,232, or 233 / berks can be 4,5,6, or 7 7 11 --> albs can be 21,22, or 23 / berks can be 44,45,46, or 47 11 7 --> albs can be 33,34, or 35 / berks can be 28,29,30, or 31 Moving on... (1) |A - B| = 7 This eliminates three of the four aforementioned possibilities for t and c, but let's take a closer look at A and B... A can be 33,34, or 35 B can be 28,29,30, or 31 Only A = 35 and B = 28 gives a difference of exactly 7. SUFFICIENT! (2) A mod 4 = 0 Without going through every possible number of Albs, both 4 and 232 are divisible evenly by 4. Not sufficient alone. ANSWER IS A

I'm guessing that "first" should say "lowest".

This is a different interpretation than I had for the problem. To me, what you're doing is keeping the order of the consonants the same...NOT the positions.

Probably because the "!" is missing.

1,000,001 1,000,010 1,000,100 1,001,000 1,010,000 1,100,000 2,000,000 Answer is B The is 7!, which is E (none of these) The answer is half of the outcomes, because half of the time it will be heads. 2^10 / 2 = 2^9 = 512 (E) There are two ways to answer this problem. 1. The easy way... Each letter is independent and has 3 possible destinations (mailboxes). Therefore, there are 3*3*3*3*3 = 3^5 = 243 possible arrangements of five letters into three mailboxes. 2. The hard way... We "count" how many ways to group the letters into three distinct piles and then multiply that by the unique ways to arrange it among the three mailboxes. 0,0,5 --> 1 group * 3 arrangements = 3 ways 0,1,4 --> 5 groups * 6 arrangements = 30 ways 0,2,3 --> 10 groups * 6 arrangements = 60 ways 1,2,2 --> 30 (10*3) groups * 3 (6/2) arrangements = 90 ways 1,1,3 --> 20 (10*2) groups * 3 (6/2) arrangements = 60 ways TOTAL = 243 ways I'm guessing that choice D is supposed to be 3^5 P(miss) = 3/4 P(miss four times) = p = (3/4)^4 = 81/256 P(not missing all four times) = 1-p = 175/256 Answer is D 1 Green Ball -> 6 ways 2 Green Balls -> 5 ways 3 Green Balls -> 4 ways 4 Green Balls -> 3 ways 5 Green Balls -> 2 ways 6 Green Balls -> 1 way 21 ways total (B) The wording is a bit off, but I think what they're saying is that spots 1,2, and 4 must NOT be the letter a. If this is the case, then it is 1/4, since there is a 1 in 4 chance that the letter a will be found randomly in spot 3. Answer is A After using an E for the first letter and an R for the last letter, we are left with the following: 1 x m 2 x e 1 x d 1 x i 1 x t 1 x r 2 x a 2 x n For the middle spaces, there are 8*7 = 56 ways permutations of two different letters from the above group, plus 3 ways to repeat them (ee, aa, and nn). Therefore, the total is 59 ways (A) There are 3 distinct ways to group the vowels together (aka in consecutive spaces): AAU AUA UAA There are 4 spaces in which to begin the group of three consecutive vowels: 1 through 4 Since each consonant is distinct, there are 3! = 6 orders possible from left to right (regardless of in which spaces they end up). Finally, we take the product: 3*4*6 = 72 ways (D)

Since the question has made no mention of which types of tables these are (ie. a round table will have a different number of distinct seating arrangements than a rectangular one) or the order of how they're seated at the table, I'm assuming that this is just a simple combinations problem. Since every distinct group of 7 people corresponds to an equally distinct group of 5, the answer is simply 12C7 = 12C5 = 12*11*10*9*8/120 = 792 The only other answer that I would consider (albeit loosely based on the vague nature of the question) would be 2*792 = 1584 (assuming each table can hold either 7 or 5 people)

Both of your observations are correct, assuming that you haven't overlooked any restrictions included in the original problems. :)

The so-called "official answers" are not reliable... QUESTION 1 Column A There are 13 distinct ranks in a standard deck of cards (assuming all jokers removed, etc). When we remove the queens, we are left with 12 ranks. So, the probability of picking a card from 2 of these 12 remaining ranks is 2/12 or 1/6. Column B There are 3 ways to roll a 4 with two standard dice: 1,3; 3,1; and 2,2 There are 4 ways to roll a 5 with two standard dice: 1,4; 4,1; 2,3; 3,2 Therefore, there are 7 ways to roll a 4 or a 5...out of 36 possible roles: 7/36 Answer is B, since 7/36 is greater than 1/6 (or 6/36) QUESTION 2 Column A 2, 3, and 5 are the only prime numbers on a standard die. 2, 4, and 6 are the only even numbers on a standard die. The total number of permutations with these restrictions is 3*3*2 - 1 = 17 {2,2 ; 2,4 ; 4,2 ; 2,6 ; 6,2 ; 3,2 ; 2,3 ; 3,4 ; 4,3 ; 3,6 ; 6,3 ; 5,2 ; 2,5 ; 5,4 ; 4,5 ; 5,6 ; 6,5} Probability is 17/36 Column B 2, 4, and 6 are the only even numbers on a standard die. 2, 4, and 6 are the only even numbers on a standard die. The total number of permutations with these restrictions is 3*3*2 - 9 = 9 {2,2 ; 2,4 ; 4,2 ; 2,6 ; 6,2 ; 4,4 ; 4,6 ; 6,4 ; 6,6} Probability is 9/36 Since there is less overlap (hence more ways to obtain it), the answer is clearly A QUESTION 3 Column A {5,6,6 ; 6,5,6; 6,6,5} => Three ways Probability = 3/216 = 1/72 Column B {1,4,4 ; 4,1,4 ; 4,4,1 ; 2,2,4 ; 2,4,2 ; 4,2,2} => Six ways Probability = 6/216 = 1/36 Answer is B QUESTION 4 Column A {1,2,3 ; 1,3,2 ; 2,1,3 ; 2,3,1 ; 3,1,2 ; 3,2,1 ; 1,3,4 ; 1,4,3 ; 3,1,4 ; 3,4,1 ; 4,1,3 ; 4,3,1} => 12 ways 12/216 = 1/18 Column B {4,4} => 1 way 1/36 Answer is A

Nicely done, Tom.

Postman, you may want to look over how to properly solve quadratic equations. Remember that they must always have a "0" on the right-hand-side... x^2 - 5x - 1 = 0 x = 5 +/- sqrt[(25+4)/2] OR x^2 - 5x + 1 = 0 x = 5 +/- sqrt[(25-4)/2]

Tom, sorry to say it but you missed the mark twice. Question 1 M2 is, indeed, 4/3, but you forgot to calculate |1-M2|, which equals 1/3, when evaluating the case for X = 0. Since 1 > 1/3, this implies that A is always greater. Question 2 Neither of your examples is valid, since each fails to take into account that n HAS TO BE less than (m+p)/2

Like question 1, question 2 requires little calculation to be solved (as is often the case with GRE stats questions)... The question tells us that n is in between m and p and that it is less than the mean of m and p. Again, going back to visualizing a number line, the mean of two numbers on a number line is the MIDPOINT of those two numbers. So here's what I immediately envision... - infinity + infinity The number of dashes is unimportant. All that matters is that n is in-between and closer to m than it is to p. Now let's look at the columns... 1. Column A Since n is less than the midpoint of m and p, we know that averaging in n will "move" the mean a little to the left (from the midpoint of m and p) but it can never equal to n itself. This is because we can represent m and p as two midpoints of m and p, which will always be greater than n. Once a mean is greater than a certain value, the only way to make that mean less than that value (in this case, n) is to average in numbers LESS THAN the value you want to equal. 2. Column B The median of m, n, and p we already know is n. Therefore, column A must always be greater than column B. A

Without doing much math at all, we can solve the first problem pretty easily I think... Question 1 Remember that standard deviation is a measure of how much the values in a group are spread out. The closer together that they are grouped, the smaller the standard deviation. If we were to plot the given data points in column B on a number line (I'm starting with the column that gives us all of the numbers), we would have two of them one unit away from 0 and one of them 2 units away from zero. Now, if we look at column A, we see that two of the three numbers are the same, so all we have to do is focus on the third. The question tells us that X is an integer less than 1, which means that it HAS TO BE FARTHER away from the other two numbers than 1. In this case, farther to the left on the number line. So, without having to even do any calculations, we can easily see that the group of points in column A -- no matter what the value of X -- will always be more spread out and will therefore have a higher standard deviation. A

Tom, you need to be careful! This problem boils down to the classic trick... sqrt(x^2) ONLY EQUALS x when x >= 0 In other words, squaring a number and then square rooting it is the same as taking its absolute value, because a square root must always be non-negative (at least in the realm of real numbers). In this problem, column B is simply column A squared and then square rooted.