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--- b:head_first_statistics:permutation_and_combination [2020/10/15 19:37] – [What if horse order doesn’t matter] hkimscil
+++ b:head_first_statistics:permutation_and_combination [2025/10/01 08:36] (current) – [exercises] hkimscil
@@ Line 3: / Line 3: @@
 ====== Permutation ======
-세마리 말이 들어오는 순서
+세마리 말이 들어오는 순서의 경우의 수
 {{:b:head_first_statistics:pasted:20191015-073815.png}}
 ===== So what if there are n horses? =====
@@ Line 72: / Line 72: @@
 b, a2, a1
-n! / p! x q!
+$$ \frac {n!} {p! * q!} $$
+<BOOKMARK:arranging_group>
 {{:b:head_first_statistics:pasted:20191015-075959.png}}
 horses
 groups 3 horses per each group
@@ Line 95: / Line 96: @@
 <WRAP box>
-__<fc #ff0000><fs xx-large>Q:</fs></fc>__ The Statsville Derby have decided to experiment with their races. They've decided to hold a race between 3 horses, 2 zebras and 5 camels, where all the animals are equally likely to finish the race first.
+X = {a a b c c c} 라면?
+n(X) = 6 이므로 총 6!
+a가 둘, c가 셋으로 묶이므로
+! / (2! * 3!)
+= 6*5*2 = 60
+</WRAP>
+<WRAP box>
+__<fc #ff0000><fs xx-large>Q:</fs></fc>__ The Statsville Derby have decided to experiment with their races. They've decided to hold a race between <fc #ff0000>3 horses, 2 zebras and 5 camels</fc>, where all the animals are equally likely to finish the race first.
 . How many ways are there of finishing the race if we’re interested in individual animals?
@@ Line 125: / Line 136: @@
 horses
 {{:b:head_first_statistics:pasted:20191015-082956.png}}
 <code>
@@ Line 144: / Line 156: @@
 {{:b:head_first_statistics:pasted:20191015-083059.png}}
+$ {}{}_{n}\mathrm{P}_{r} $
 ===== What if horse order doesn’t matter =====
@@ Line 160: / Line 173: @@
 \end{eqnarray*}
+Among the two, the order doesn't matter. Then, we follow the same logic as [[#arranging_group|the above]]
 representatives
 A B | B A
@@ Line 167: / Line 180: @@
 \begin{eqnarray*}
-_{3}C_{2} * 2! & = & _{3}P_{2} \\
+\text{Answer we want} & = & \frac {_{3}P_{2}}{2!} \\
-_{3}C_{2} & = & \frac {_{3}P_{2}}{2!} \\
+\text{We call this} & = &  _{3}C_{2}  \\
 _{3}C_{2} & = & \frac {\frac{3!}{(3-2)!}} {\frac {2!} {1}} \\
 _{3}C_{2} & = & \frac {3!}{2! * (3-2)!} = 3
@@ Line 189: / Line 202: @@
 {{:b:head_first_statistics:pasted:20191015-084051.png}}
-$\displaystyle {^{n} P_{r}} $
+\begin{eqnarray*}
-$\displaystyle ^{n} P_{r} = \displaystyle \frac {n!} {(n-r)!}$
+\displaystyle ^{n} P_{r} = \displaystyle \dfrac {n!} {(n-r)!} \\
-A permutation is the number of ways in which you can choose objects from a pool, and where the order in which you choose them counts. It’s a lot more specific than a combination as you want to count the number of ways in which you fill each position.
+\end{eqnarray*}
+A **permutation** is the number of ways in which you can choose objects from a pool, and **where the order in which you choose them counts**. It’s a lot more specific than a combination as you want to count the number of ways in which you fill each position.
-$\displaystyle ^{n} C_{r}$
+\begin{eqnarray*}
-$\displaystyle ^{n} C_{r} = \displaystyle \frac {n!} {r! \cdot (n-r)!}$
+\displaystyle ^{n} C_{r} & = & \displaystyle \dfrac {^{n} P_{r}} {r!} \\
-A combination is the number of ways in which you can choose objects from a pool, without caring about the exact order in which you choose them. It’s a lot more general than a permutation as you don’t need to know how each position has been filled. It’s enough to know which objects have been chosen.
+& = & \displaystyle \frac {n!} {r! \cdot (n-r)!} \\
+\end{eqnarray*}
+A **combination** is the number of ways in which you can choose objects from a pool, **without caring about the exact order in which you choose them**. It’s a lot more general than a permutation as you don’t need to know how each position has been filled. It’s enough to know which objects have been chosen.
 ===== e.g. =====
@@ Line 206: / Line 222: @@
 . The coach classes 3 of the players as expert shooters. What’s the probability that all 3 of these players will be on the court at the same time, if they’re chosen at random?
 </WRAP>
+<WRAP box>
+$ {}_{52} P _{5} $
+<code>
+# only combination function is available in r, choose
+# for permutation
+> choose(52,5)
+[1] 2598960
+> permute <- function(n,r) {
+>   choose(n,r) * factorial(r)
+> }
+> permute(52, 5)
+> [1] 311875200
+> # or
+> factorial(52)/factorial(52-5)
+[1] 311875200
+>
+</code>
+</WRAP>
+답. 12명 중에서 순서는 상관없는 5명이므로
+${}_{12} C _{5} $
 <code>
 ## n! / r!(n-r)!
@@ Line 217: / Line 254: @@
 a
 b
-b/a
 </code>
 <code>
@@ Line 230: / Line 266: @@
 > b
 [1] 36
-> b/a
-[1] 0.04545455
 >
 </code>
@@ Line 244: / Line 278: @@
 A flush is where all 5 cards belong to the same suit. What’s the probability of getting this?
 </WRAP>
+{{https://upload.wikimedia.org/wikipedia/commons/thumb/0/02/Piatnikcards.jpg/1920px-Piatnikcards.jpg?800}}
+see [[wp>List_of_poker_hands]]
+{{https://upload.wikimedia.org/wikipedia/commons/thumb/e/e2/A_studio_image_of_a_hand_of_playing_cards._MOD_45148377.jpg/1024px-A_studio_image_of_a_hand_of_playing_cards._MOD_45148377.jpg?400}}
 <code>
 ## 52장의 카드 중에서 5장 고를 조합은
 factorial(52)/(factorial(5)*factorial(52-5))
-all <- factorial(52)/(factorial(5)*factorial(52-5))
+all2 <- factorial(52)/(factorial(5)*factorial(52-5))
+all2
+# or
+all <- choose(52, 5)
+all
 ## royal flush = 10, 11, 12, 13, 1 각 문양
 ## 즉, 4가지. 따라서 전체 조합 중 4가지만 해당됨
@@ Line 299: / Line 339: @@
 </code>
+===== exercises =====
+말 3마리, 얼룩말 4마리, 그리고 낙타 6마리 중에서 낙타 4마리와 얼룩말 3마리를 뽑아서 일렬로 앉히는 경우의 수를 구하시오.
+<code>
+> # 6C4 * 4C3 * 7!
+> choose(6,4) * choose(4,3) * factorial(4+3)
+[1] 311875200
+>
+</code>
+tennessee 를 서로 다른 방법으로 배치할 수 있는 경우의 수는?
+<code>
+> # t
+> # eeee
+> # nn
+> # ss
+> # 9 letters
+> # 9!/4!*2!*2!
+> factorial(9)/(factorial(4)*factorial(2)*factorial(2))
+[1] 3780
+>
+</code>
+AGAIN이라는 단어가 있다. 이 단어를 알파벳 순으로 조합하여 나열한다면 49번째 오는 단어는 어떤 것일까?
+처음 A 로 시작하게 되는 단어는 G A I N 의 단어를 조합하여 나오는 수 4! = 24 단어
+다음에는 G 로 시작하는 단어 A A I N 의 단어의 조합은 4!/2! = 12 단어
+I 로 시작하는 단어는 A G A N 의 조합은 4!/2! = 12 단어
+그렇다면 N으로 시작하는 첫단어가 답
+N A A G I
+S M I L E 이라는 단어의 문자에서 3 개를 뽑아서 나열하는 경우의 수는?
+$ _{5}P_{3} = \dfrac {5!}{2!} = 60 $
+AJOU UNIVERSITY 단어에서 AJOU 네 단어가 서로 이웃해서 모든 단어가 나열되는 경우의 수는?
+X U N I V E R S I  T  Y
+2 3 4 5 6 7 8 9 10 11
+글자의 조합은 11! / 2!
+A J O U 의 조합도 신경을 써야 하므로 4! 을 곱해준다.
+POWERFUL 라는 단어의 글자들을 나열하려고 한다. 모음이 앞이나 뒤에 적어도 한번은 들어가도록 나열하는 경우의 수는? W는 자음
+이다.
+모 . . . . 모
+모 . . . . 자
+자 . . . . 모
+자 . . . . 자
+의 경우라고 생각해야 할 듯
+모음은 O E U
+자음은 P W R F L
+전체 글자는 8 글자   8!
+양쪽에 자음이 오는 경우는 5P2 = 20
+\begin{eqnarray*}
+{n \choose x} \\
+\binom{n}{r} \\
+_{n}C_{r} \\
+^{n}P_{r} \\
+_{n}P_{r} \\
+\end{eqnarray*}