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

Q1.

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?

A.

2*1*2*1 = 4

Q2.

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?

A.

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

Q3.

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?

A.

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

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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
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(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.

 ACBD  ACDB  BACD  DACB  BDAC  DBAC
 CABD  CADB  BCAD  DCAB  BDCA  DBCA

(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.

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

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