Answer: Hello!
Let's start with the word TRISKAIDEKAPHOBIA wich has 17 letters (some of them repeat, but it does not matter in this problem)
We want to know how many permutations we can do with 17 letters: then think this way, Lets compose a word. The first letter of this word has 17 options, the second letter of the word has 16 options (you already took one of the set) the third letter of the word has 15 options, and so on.
The total number of permutations is the product of the number of options that you have for each letter, this is:
17*16*15*14*....*3*2*1 = 17! = 3.6e+14
(b) FLOCCINAUCINIHILIPILIFICATION now we have 30 letters in total, using the same reasoning as before, here we have 30! permutations; this is
30! = 2.65e+32
(c) PNEUMONOULTRAMICROSCOPICSILICOVOLCANOCONIOSIS: now there are 47 letters.
then P = 47! = 2.59e+59
(d) DERMATOGLYPHICS: here are 18 letters, then:
p = 18! = 6.4e+15
Answer:
2/3
Step-by-step explanation:
Answer:
1/6
1/12
1/2
Step-by-step explanation:
There are three possible outcomes in the left spinner, and four possible outcomes in the right spinner. So there are a total of 3×4=12 possible combinations. We can show that by making a grid:
![\left[\begin{array}{cccc}&R&B&G\\R&RR&BR&GR\\B&BR&BB&BG\\P&PR&BP&GP\\Y&RY&BY&GY\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26R%26B%26G%5C%5CR%26RR%26BR%26GR%5C%5CB%26BR%26BB%26BG%5C%5CP%26PR%26BP%26GP%5C%5CY%26RY%26BY%26GY%5Cend%7Barray%7D%5Cright%5D)
Of these 12 combinations, 2 show both spinners landing on the same color (RR and BB). So the probability is 2/12 = 1/6.
There is only 1 outcome in which the first spinner lands on R <em>and</em> the second spinner lands on P (PR), so the probability is 1/12.
There are 6 outcomes in which the first spinner lands on R <em>or</em> the second spinner lands on P (RR, BR, PR, RY, BP, GP). So the probability is 6/12 = 1/2.
Answer:
dont know
Step-by-step explanation:
dont know
It looks like it’s already answered I’m confused about that
