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--- b:head_first_statistics:permutation_and_combination [2025/09/30 15:31] – [exercises] hkimscil
+++ b:head_first_statistics:permutation_and_combination [2025/09/30 23:36] (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 136: / Line 136: @@
 horses
 {{:b:head_first_statistics:pasted:20191015-082956.png}}
 <code>
@@ Line 155: / Line 156: @@
 {{:b:head_first_statistics:pasted:20191015-083059.png}}
+$ {}{}_{n}\mathrm{P}_{r} $
 ===== What if horse order doesn’t matter =====
@@ Line 200: / 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 219: / Line 224: @@
 <WRAP box>
+$ {}_{52} P _{5} $
 <code>
 # only combination function is available in r, choose
@@ Line 224: / Line 230: @@
 > choose(52,5)
 [1] 2598960
-> perm <- function(n,r) { choose(n,r)*factorial(r)}
+> permute <- function(n,r) {
-> perm(52, 5)
+>   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 242: / Line 254: @@
 a
 b
-b/a
 </code>
 <code>
@@ Line 255: / Line 266: @@
 > b
 [1] 36
-> b/a
-[1] 0.04545455
 >
 </code>
@@ Line 334: / Line 343: @@
 <code>
 > # 6C4 * 4C3 * 7!
-> choose(6,4) * choose(4,3) * (4+3)
+> choose(6,4) * choose(4,3) * factorial(4+3)
-[1] 420
+[1] 311875200
 >
 </code>
@@ Line 360: / Line 369: @@
 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