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--- b:head_first_statistics:permutation_and_combination [2023/10/11 08:15] – [e.g.] hkimscil
+++ b:head_first_statistics:permutation_and_combination [2024/10/01 22:40] (current) – [e.g.] hkimscil
@@ Line 106: / Line 106: @@
 <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.
+__<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 217: / Line 217: @@
 . 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>
+<code>
+# only combination function is available in r, choose
+# for permutation
+> choose(52,5)
+[1] 2598960
+> perm <- function(n,r) { choose(n,r)*factorial(r)}
+> perm(52, 5)
+> [1] 311875200
+</code>
+</WRAP>
 <code>
@@ Line 258: / Line 271: @@
 {{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}}]
+{{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