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

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

astha8579 astha8579 · Answer 2 · 2022-02-14T12:11:23+01:00

You can use the fact that n distinct items can be written n! ways.

The permutations needed are

a) [tex]\dfrac{17!}{2!}[/tex]

b) [tex]\dfrac{29!}{3! \times 2! \times 2! \times 3! \times 4! \times 2!}[/tex]

c) [tex]\dfrac{46!}{2! \times 3! \times 4! \times 2! \times 6! \times 6! \times 4! \times 2! \times 2! \times 2!}[/tex]

d) [tex]\dfrac{17!}{2!}[/tex]

What is the number of permutations in which n things can be arranged such that some groups are identical?

Suppose there are n items.

Suppose we have [tex]i_1, i_2, .., i_k[/tex] sized groups of identical items.

Then the permutations of their arrangements is given as

[tex]\dfrac{n!}{i_1! \times i_2! \times ...\times i_k!}[/tex]

Using the above conclusion, we get:

a) TRISKAIDEKAPHOBIA

There are n = 17 letters

The frequency is

{'S': 1, 'K': 2, 'P': 1, 'A': 3, 'D': 1, 'I': 3, 'R': 1, 'E': 1, 'O': 1, 'B': 1, 'T': 1, 'H': 1}

The number of permutations of this word's arrangement is

[tex]\dfrac{17!}{2!}[/tex]

b) FLOCCINAUCINIHILIPILIFICATION

There are n = 29, and the frequency of letters is
{'P': 1, 'U': 1, 'N': 3, 'F': 2, 'T': 1, 'A': 2, 'L': 3, 'C': 4, 'I': 9, 'O': 2, 'H': 1}

The number of permutations of this word's arrangement is

[tex]\dfrac{29!}{3! \times 2! \times 2! \times 3! \times 4! \times 2!}[/tex]

c) PNEUMONOULTRAMICROSCOPICSILICOVOLCANOCONIOSIS

There are 46 letters and the frequency of letters is

{'P': 2, 'L': 3, 'N': 4, 'U': 2, 'I': 6, 'E': 1, 'T': 1, 'C': 6, 'S': 4, 'V': 1, 'A': 2, 'M': 2, 'R': 2, 'O': 9, ' ': 1}

The number of permutations of this word's arrangement is

[tex]\dfrac{46!}{2! \times 3! \times 4! \times 2! \times 6! \times 6! \times 4! \times 2! \times 2! \times 2!}[/tex]

d) DERMATOGLYPHICS

There are 17 letters and the frequency of letters is

{'S': 1, 'I': 1, 'Y': 1, 'H': 1, 'G': 1, 'O': 1, 'A': 1, 'R': 1, 'L': 1, 'D': 1, 'T': 1, 'C': 1, 'M': 1, ' ': 2, 'P': 1, 'E': 1}

The number of permutations of this word's arrangement is

[tex]\dfrac{17!}{2!}[/tex]

Thus,

The permutations needed are

a) [tex]\dfrac{17!}{2!}[/tex]

b) [tex]\dfrac{29!}{3! \times 2! \times 2! \times 3! \times 4! \times 2!}[/tex]

c) [tex]\dfrac{46!}{2! \times 3! \times 4! \times 2! \times 6! \times 6! \times 4! \times 2! \times 2! \times 2!}[/tex]

d) [tex]\dfrac{17!}{2!}[/tex]

Learn more about permutations here:

https://brainly.com/question/6743939