Permutations and Combinations only for geniuses?

[Fakesmile]

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Am I dumb or what? I'm taking a probability course and it's covering permutations and combinations right now. I can do the easy questions but never the harder ones like:

Three married couples purchased plane tickets and are sitted in a row with just six seats. If they take their seats in a random order, what is the probability that Bob and Jane (husband and wife) sit in the two seats on the far left? Next to one another? What is the probability that at least one of the wives ends up sitting next to her husband?

I looked at the answer key and the answer to the question above involved some crazy multiplications of several different combinations. It just doesn't make sense.

I can't seem to be able to do these kinds of problems no matter how hard I try (though I can do all the basic easy problems). Do you think it takes an innate mathematical brain to master combinations and permutations?

 

[Podalarius]

Prob and stats can be counterintuitive at first, but it's very helpful to break it down to simpler units, and to sometimes just draw things out. It has been about eight years since I took prob and stats, but:

If Bob and Jane are the two seats in the far left...
There are two options for the leftmost seat, and the other one fills the second seat. The remainder can sit in any order. 2 * 1 * 4!. There are 6! arrangements of 6 unique people in 6 chairs, so the probability is 2 * 1 * 4!/6!

If Bob and Jane are next to each other...
Imagine that Bob and Jane are in fact one unit, and there are five chairs. Since any seating arrangement treating them as one unit satisfies the conditions of the problem, there are 5! (any arrangement of seats) * 2 (for Bob/Jane and Jane/Bob). There are 6! arrangements of 6 unique people in 6 chairs, so the probability is (5!*2)/6! = 1/3

Alternatively, if one of them is sitting in an edge chair, there is one valid option, while there are two options in any of the center chairs. Since there are 6 outcomes for both Bob and Jane (and this method counts each of them twice) there are 2/2 * 1/6 (1/5 remaining seats * 2 edge seats + 2/5 remaining * 4 center seats) = 1/3

Any husband and wife sit next to each other...
Like above, the probability that a given couple is sitting together is (5!*2)/6!. While it would be tempting to multiply that probability by three to get the total, that's not an option, because it counts some instances of couples sitting next together twice. This is easier to picture with a Venn diagram, with the outcome of a given couple sitting together intersecting with the probability of 2 or 3 couples sitting together. The total for each couple must be the probability calculated above. The probability of each couple sitting together is essentially the same as 3 people rotating in 3 chairs or 3!, but each of the three couples exists in 2 states, so it is actually 3! * 2^3. The probability of only two couples sitting next together is given by the same logic: 4! * 2^2, but, of those, 6 are actually instances where all three are next to each other, so 4! * 2^2 - 6 There are 3 combinations of 2 couples, so this happens three times (i.e. at each intersection of two circles of a Venn diagram). The probability of a single couple sitting together while the other two do not is therefore 5! *2 /6! - 3! * 2^3 - 4! *2^2 + 6, and the probability of at least one couple sitting together is therefore 3*(5!*2/6!) - 3! * 2^3 - 3* (4! * 2^2 - 6).

I have no idea if any of this is right, but that's how I'd try to work on that problem. Break it into smaller pieces, draw out what you know, and then solve for each piece.

 

[Fakesmile]

What you wrote down is different from what's on the answer key, but there is more than one right way of tackling these kinds of problems and your approach seems reasonable and on the right track.
But looking at your explanations, I think my head is going to explode.

 

[mirrorpair]

You don't learn math, you just get used to it.

 

[Jolie South]

SDN is not the place to seek help with homework. Closing this thread.