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Permutation and combination 唔明個answer

Permutation and combination 唔明個answer

Question:
9 (a) 10 students are arranged in 2 rows of 5 in order to take a photograph. Find the number of arrangements that can be made of
(ii) two particular students are next to each other

Answer:
2 X  < (8C5 x 5!) (3C3 x 3!) x 4 X 2)>  = 645,120
唔明個answer 點尼
但我用另一個方法
2!9!-8!2! 我又計到個answer 係岩,點解?

[ 本帖最後由 zxc123zxc 於 2021-9-20 10:24 PM 編輯 ]
   

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[隱藏]
I think a good gut check with combinatorics (i.e. combination, permutation, probability) is that you can describe in words what each number and each symbol mean. Sidenote: Symbols are easy: "+" means "or", "x" means "and", "-" means "but not".

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Pre-word 1: Not sure why the 1st method uses nCr, when it's an arrange stuff in order problem.
Pre-word 2: I remember this problem having some ambiguity. Do the 2 students next to each other have to be in the same row? Why can't they are in different rows, just back and front? But anyway, we all know what it supposed to mean, so... er... yeah... just me being the usual pedantic math guy.

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1st Method:
8C5: First ignore the 2 particular students, pick any 5 students from the remaining 8*;
x5!: AND arrange them in a row, say the back row for now*;
x3C3: AND still ignoring the 2 particular students, pick any 3 students from the remaining 3**;
x3!: AND arrange them in a row, front row for now**;
x4: AND there are 4 positions to put the pair of students in the front row after the relative positions of the 3 are determined**;
x2: AND 2 ways to arrange the pair**;
x2 (at the front): AND swapping front and back row counts as different arrangements.

*I think it's needlessly clumsy. It's just 8P5 - an example problem in your textbook (or as I prefer, 8x7x6x5x4. I think nPr is unintuitive in general).

**Again, needlessly clumsy. Simply put, it's the same idea as putting 3 English books and 2 Maths books on a shelf, with the 2 Maths books being together. What we tend to do is just 4!x2! - another example problem in your textbook.

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I think I'm gonna leave the explanation of your method to you as an exercise, i.e. I don't have an immediate answer for you now. Feel free to post it here for sharing.

P.S. Often times there are different ways to solve the same problem. As long as your method makes sense, you can use it.

[ 本帖最後由 風之男 於 2021-9-20 08:34 PM 編輯 ]

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成題係道,消化下先。

[ 本帖最後由 zxc123zxc 於 2021-9-20 10:24 PM 編輯 ]
附件: 您所在的用戶組無法下載或查看附件

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https://www.youtube.com/watch?v= ... vNiJEs&index=91

第2個方法係用herman Yeung教既方法做。

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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 ]

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