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7 people are attending a concert.

(a) In how many different ways can they be seated in a row?

(b) Two attendees are Alice and Bob. What is the probability that Alice sits next to Bob?

(c) Bob decides to make Alice a rainbow necklace with 7 beads, each painted a different colour on one side (red, orange, yellow, blue, green, indigo, violet), placed on a chain that is then closed to form a circle. How many different necklaces can he make? (Since the beads can slide along the chain, the necklace with beads R O Y G B I V would be considered the same as O Y G B I V R for example. The beads are plain on the back, so the necklace cannot be turned over.)

How should i approach these questions? Are they correct?

for the first one i understand that it is a permutation. Therefore would a) = 7! = 5040 possible different ways of sitting in a row

b) p(7,2) = 7!/(7-2)! = 5040/120 = 42 therefore probabiltiy = 42/5040 = 0.0083%

c) =6!/2 because the first bead doesnt matter and over 2 as it can either go left or right.

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    $\begingroup$ The number $7!$ is right. Your derivation is unnecessarily long. $\endgroup$ – André Nicolas Sep 25 '14 at 1:12
  • $\begingroup$ Is the answer simply 7! ? 5040 different ways to be seated in a row sounds a bit excessive for me. $\endgroup$ – I hate math Sep 25 '14 at 1:14
  • $\begingroup$ The problem is unfortunately not fully specified. If we had been told that there are say $12$ seats in the row, we would be expected to take that into account. As it is, we are tacitly asked to only worry about the relative positions of the eople. $\endgroup$ – André Nicolas Sep 25 '14 at 1:17
  • $\begingroup$ Is my question b) correct? $\endgroup$ – I hate math Sep 25 '14 at 4:14
  • $\begingroup$ As a probability question it should also have stipulated that all possible seatings are equally likely. Since Bob apparently likes Alice, that may not be a realistic assumption. $\endgroup$ – Marc van Leeuwen Sep 25 '14 at 4:45
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For Question (b), we want to count the number of seatings in which Alice and Bob are neighbours. We are assuming the seating is random. That may not be reasonable, if we consider Bob's actions described in (c).

The leftmost of the two seats occupied by our two heroes can be chosen in $6$ ways. For each such way, Alice and Bob can occupy that seat and the next one in $2$ ways. And then the rest of the people can fill in the rest of the spots in $5!$ ways, for a total of $(6)(2)(5!)$.

For the probability, divide by $7!$. Simplify. Fairly quickly we arrive at $\frac{2}{7}$.

Another way: There are $\binom{7}{2}$ equally likely ways to choose a set of two seats. Of these, $6$ sets have Alice and Bob next to each other. So the required probability is $\frac{6}{\binom{7}{2}}$. This simplifies to $\frac{2}{7}$.

Still another way: The probability that Alice occupies an end seat is $\frac{2}{7}$. If she does, the probability Bob is next to her is $\frac{1}{6}$.

The probability that Alice occupies a non-end seat is $\frac{5}{7}$. If she does, the probability Bob is next to her is $\frac{2}{6}$. Thus the required probability is $\frac{2}{7}\cdot\frac{1}{6}+\frac{5}{7}\cdot\frac{2}{6}$. This simplifies to $\frac{2}{7}$.

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Since André Nicolas only left (c) to be answered I'll just mention that one. Since no starting point for the beads is fixed, one can choose any one bead as reference point and slide it to the first position; I'll arbitrarily choose the orange bead for this (O looks a bit like $0$). Then each necklace is uniquele determined by "seating" the remaining beads Y G B I V R in the six remaining positions relative to O; this is like problem (a) but with $6$ in place of $7$.

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