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Some questions

This topic has 1 member reply
maxim730 Senior | Next Rank: 100 Posts Default Avatar
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Some questions

Post Sun Jan 14, 2007 11:49 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    I would appreciate if someone can explain the answers to me Smile

    Q: (Source: Princeton Review Cracking the GMAT online exam)

    Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

    A: 1/5, 1/4, 3/8, 2/5, or 1/2

    OA: 2/5

    PR Explanation: Yes. Let's call the first couple C and c, the second couple K and k, and the single person S. Let's seat S in different places and figure out the possible ways to have no couples sit together. If S sits in the first seat, any of the remaining four people could sit next to S. However, only two people could sit in the next seat: the two who don't form a couple with the person just seated). For example, if we have S K so far, C or c must sit in the third seat. Similarly, we have only one choice for the fourth seat: the remaining person who does not form a couple with the person in the third seat. Because we have seated four people already, there is only one choice for the fifth seat; the number of ways is 4 2 1 1 = 8. Because of symmetry, there are also 8 ways if S sits in the fifth seat. Now let's put S in the second seat. Any of the remaining four could sit in the first seat. It may appear that any of the remaining three could sit in the third seat, but we have to be careful not to leave a couple for seats four and five. For example, if we have C S c so far, K and k must sit together, which we don't want. So there are only two possibilities for the third seat. As above, there is only one choice each for the fourth and fifth seats. Therefore, the number of ways is 4 2 1 1 = 8. Because of symmetry, there are also 8 ways if S sits in the fourth seat. This brings us to S in the third seat. Any of the remaining four can sit in the first seat. Two people could sit in the second seat (again, the two who don't form a couple with the person in the first seat). Once we get to the fourth seat, there are no restrictions. We have two choices for the fourth seat and one choice remaining for the fifth seat. Therefore, the number of ways is 4 2 2 1 = 16. We have found a total of 8 + 8 + 8 + 8 + 16 = 48 ways to seat the five people with no couples together; there is an overall total of 5! = 120 ways to seat the five people, so the probability is = 48/120 = 2/5. Whew!


    Q: Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five models are to stand in a line from shortest to tallest, and the fourth-tallest and sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?

    A: 6,11,17,72, 210

    OA: 72

    PR Explanation: No. Forget about the fourth-tallest and sixth-tallest jazz and start by finding the number of ways five models could be arranged. The tricky part here is that, because the models must stand in a particular order, the order in which they are chosen does not matter. Therefore, this is a combination problem; 7!/5!(2!)= 21 ways. We can eliminate (D) and (E) because they're too big. Now it's time to abandon the formulas. If we try to list some different ways that the fourth-tallest and sixth-tallest models could be adjacent, there are only four: 12346, 12467, 13467, and 23467. That leaves 17 possible ways to arrange the models. The correct answer is (C).

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    thankont Senior | Next Rank: 100 Posts
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    Post Mon Jan 15, 2007 12:02 am
    Q2)
    My ans. is 17
    Look at 3 first
    --3--
    we have 5 cases 12345, 12347, 12356, 12357, 12367
    -3---
    first place could be 1 or 2 while last 3 places could be 456 or 457 or 567
    so 2x3=6 cases
    3---- only with 4567 1case

    now take 3 out of the picture and look at 4
    --4--
    last 2 digits could be 56 or 57 so 2 cases
    -4---
    first place could be 1,2 last digits 567 so 2 cases

    finally 3, 4 out so we have 1 case 12567

    total = 5+6+1+2+2+1 = 17
    --like to see different approaches--
    thanx

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