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GREYUIT [131]
3 years ago
6

How many permutations are there of the letters in the words (a) TRISKAIDEKAPHOBIA (fear of the number 13)? (b) FLOCCINAUCINIHILI

PILIFICATION (estimating something as worthless)? (c) PNEUMONOULTRAMICROSCOPICSILICOVOLCANOCONIOSIS (a lung disease caused by inhaling fine particles of silica)? (This word is, by some accounts, the longest word in the English language.) (d) DERMATOGLYPHICS (skin patterns or the study of them)? (This word is the (current) longest word in the English language that doesn't repeat a letter; another word of the same length is UNCOPYRIGHTABLE.13)
Mathematics
1 answer:
zubka84 [21]3 years ago
6 0

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

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Step-by-step explanation:

1. Change the 5, the denominator to 10.  8/10

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2. Change both denimnators to 15. You will then get 9/15 and 5/15.

Add them together and get 14/15

3. Change both denominators to 6 and you will get 3/6 and 4/6.

Add them together to get 7/6.

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