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小 發表於 2021-9-21 12:06 AM (第 920 天)
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I've given some thoughts on Method 2. The big idea is to basically ignore the fact that it's 2 separate rows, but treat it as one row first, then just take away those that are not right.
First, you want to arrange 10 folks in order, with 2 friends standing next to each other. That's easy - 9!x2!. Then we can just split the queue right in the middle to form 2 separate rows. In case you have worries, you won't double count any arrangements here (beware, this is not easy to understand for first-timers. Believe me, I struggled too back then). However, rather than trusting me (and Herman, probably), try to think why is that the case (that's your exercise now, mostly because I'm lazy).
Now, there is one scenario which doesn't fit our requirement, which is the following:
_ _ _ _ A | B _ _ _ _
This particular arrangement has 8!x2! ways.
Hence, 9!x2! - 8!x2!. (In other words, arranging 10 folks in a queue with 2 friends standing next to each other, BUT NOT when the 2 friends are standing in positions 5 and 6.)
I think the first method is easier to understand (barring the super clumsy way of presenting it), but the second method looks way cooler. Of course, under time pressure of exams, we don't really care about elegance. But then I wish you could appreciate how cool math is - it's not as stiff as most people think (a.k.a. the "appreciate math" part in the curriculum, which no one cares).
[ 本帖最後由 風之男 於 2021-9-21 12:08 AM 編輯 ]