排列組合 - 直線排列 Permutation and Combination - Permutation in Linear Order

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排列組合 - 直線排列

Permutation and Combination - Permutation in Linear Order

第 1 節 1st Period

Material Note

Word：Permutation (排列), Tree diagram (數狀圖), Arrangement (安排的情況), Scenario (情況).

Sentence：

1. If we seat them in order, how many different people could sit in the 1^st seat? (如果將他們按順序排成一列，第一個位子有幾個人可入座？) 2. For each of the 3 scenarios, after the first seat is

taken, 2 different people could be put in the 2^nd seat. (在這 3 種情況下，第 2 張椅子有 2 個人可以坐。)

3. We can draw a tree diagram, the first section has 3 branches with person A, B, or C. (使用樹狀圖的話，第一個分支即有 3 個人 A，B 或 C。) Word：Factorial (階乘)

Sentence：

The number of permutations for seating these n people in n steats is n times n-1 times n-2 times to 1, which we call it n factorial. (n 個人坐 n 張椅子的排列數是，n 乘 n-1 乘 n-2 乘到 1，我們稱為「n 的階乘」。)

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Translation:

In this question, we have 6 people needed to fit in 6 places. So the permutations for this question is 6 factorial, which is 6 times 5 times 4 times 3 times 2 times 1. And it is equal to 720 arrangements.

Sentence:

1. How can I relate factorial to this problem? (要如何連結至階乘呢？)

2. It looks like we kind of did factorial, but we stopped. We didn’t go times 4 times 3 times 2 times 1. (可以用另一種想法，我們其實是使用了 7 的階乘，但沒有乘 4 到 1。)

3. We can write it in terms of factorial. We could write this as 7 factorial over 4 factorial. (我們就可以用階乘把算式寫成 7 的階乘除以 4 的階乘。) Suggested Instruction:

In conclusion, we have a notation P_kⁿ for the number of permutations where we put n people in k chairs is going to be n factorial over n minus k

factorial.

Note:

n

Pk can be written inⁿP_k, _{n k}P , nPk or ^{P n k}

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Translation:

In this question, there are 7 songs, but there are only 4 shows to perform. Therefore, we have 7 times 6 times 5 possible scenarios to give a performance, which is equal to P₄⁷, also is 210.

補充題 Material

Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that

(i) All vowels occur together (ii) No vowels occur together Solution:

(i) There are 8 different letters in the word DAUGHTER. There are 3 vowels, namely, A, U and E.

Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be 6P6 = 6!. Corresponding to each of these permutations, we shall have 3!

permutations of the three vowels A, U, E taken all at a time. Hence, by the multiplication principle the required number of permutations = 6 ! × 3 ! = 4320.

(ii) If we have to count those permutations in which no vowels can be together, we first have to find all possible arrangements of 8 letters taken all at a time, which can be done in 8! ways.

Then, we have to subtract from this number, the number of permutations in which the vowels are always together. Therefore, the required number 8 ! – 6 ! × 3 ! = 6 ! (7×8 – 6) = 2 × 6 ! (28 – 3) = 50 × 6 ! = 50 × 720 = 36000

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Note Word: Vowel (母音).

Sentence:

1. All vowels occur together. (母音完全相鄰) 2. No vowels occur together. (母音不完全相鄰)

3. Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE). (因為要將母音排在一起，我們可以將 AUE 先視為一體。)

4. We count permutations of these 6 objects taken all at a time. (我們先數 6 個物品的排列數。)

參考資料 References

1. 許志農、黃森山、陳清風、廖森游、董涵冬（2019）。數學 2：單元 4 排列。龍騰文化。

2. National Council of educational Research & Training. (2022, April 10). Permutation and Combinations FINAL 04.01.PMD. https://ncert.nic.in/textbook/pdf/kemh107.pdf.

3. Khan Academy. (2022, April 10). Unit: Counting, permutations, and combinations.

https://www.khanacademy.org/math/statistics-probability/counting-permutations-and- combinations.

製作者：臺北市立陽明高中吳柏萱教師