I'm practicing some GRE math problems which I don't have answers for. Would you please correct my answers if they're wrong?


You have 4 books, A and B and C and D. In how many ways you can arrange them on a shelf and keep book A and book C next to each other?


2*1*2*1 = 4


Anne closet contain 11 pants, 4 of them are blue. 7 hats, 3 of them are blue. 8 shirts, 2 of them are blue. What's the probability that Anne would grab a blue pants, a blue hat and a blue shirt?


1/4 * 1/3 * 1/2 = 1/24


A set of 7 different integers, it's range is 12, and it's median is 86. What's the lease possible of the first integer in the set?


integers after median in the set would be: {,,,86,87,88,89} and 89-12=77 So answer is 77.

  • 2
    $\begingroup$ The answer to Q1 is not right. Imagine book A and book C as one big book F. In how many ways you can arrange B,D and F ? Additionally you have to consider, that A and C can be arranged in two ways. $\endgroup$ Aug 13 '15 at 5:55
  • 1
    $\begingroup$ For Q2, consider the probability that she picked a blue pair of pants ($\frac{4}{11}$), the probability that she picked a blue hat ($\frac{3}{7}$), and the probability she picked a blue shirt ($\frac{2}{8}$). It seems safe to assume a uniform probability distribution for each and that they are independent, so to find the probability of all happening, you multiply the probabilities for $\frac{4}{11}\cdot\frac{3}{7}\cdot\frac{2}{8}$. (I had a typo in my earlier comment, caught by @calculus) $\endgroup$
    – JMoravitz
    Aug 13 '15 at 6:10
  • $\begingroup$ @calculus i see. so it would be 3*2*1*2 ? or 3*2*1? Thanks! $\endgroup$
    – Shadin
    Aug 13 '15 at 6:17
  • $\begingroup$ @JMoravitz Many thanks! $\endgroup$
    – Shadin
    Aug 13 '15 at 6:18
  • 1
    $\begingroup$ @Shadin The first one is right. B,D,F can be arranged in $3!=1*2*3=6$ ways. And A and C can be arranged in 2 ways. $\endgroup$ Aug 13 '15 at 6:19

(1) Consider 'AC' as a single object. There are 3! ways in which to arrange three objects 'AC', B and D. Then multiply by 2 because A and C are still next to each other if it's 'CA'. Answer: 12 ways.


(2) Answered in Comment by @JMoravitz

(3) "77, two intermediate digits, 86, two intermediate digits, 89" has 7 digits, median 86, range 89 - 77 = 12, and s0 satisfies the requirements.

But if the integers didn't need to be DIFFERENT, we could have "75, 77, 79, 86, 87, 87, 87" for a smaller minimum number.

  • $\begingroup$ Wishing you a magnificent score on the GRE. $\endgroup$
    – BruceET
    Aug 13 '15 at 8:38

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