Question 1082067
IF LETTERS CAN BE REPEATED,
AAAAA could be considered a word.
Then, a lot more words could be formed.
(i) You have 5 options for each of the 5 letters,
so you can form {{{5*5*5*5*5=5^5=3125}}} 5-letter words.
(ii) There would be the 4 consonants,
arranged in any of {{{4!=4*3*2=24}}} different orders,
plus an additional letter.
Of the additional letter chosen is a vowel,
there are 5 letter choices and 5 position choices.
That would yield {{{5*5*4*3*2=600}}} words including all consonants and a vowel.
Choosing one of the 4 consonants as the additional letter,
would yield {{{4*5*4*3*2}}} different arrangements IF
we could distinguish the added consonsnt
from the one just like it in the 4 arranged first.
Since they cannot be distinguished,
the number of different arrangements of 5 vowels formed would be only
{{{4*5*4*3=240}}} .
So the number of 5-letter words (repetition allowed)
containing all 4 consonants would be
{{{600+240=840}}} .
(iii) If the 5-letter word must begin with a certain letter,
and send with another letter,
there are {{{9^3=729}}} arrangements of all 9 letters
that could be used to fill the 3 positions in between.
 
 
IF LETTERS CANNOT BE REPEATED:
(i) There are {{{5!=5*4*3*2=120}}} permutations of 5 different vowels.
 
(ii) There are only {{{4}}} consonants in the word EQUATIONS, so to make a 5-letter word, with all 4 consonants, and no repeated letters, a vowel would have to be included .
With {{{4!=4󫢪=24}}} possible permutations of the consonants,
{{{5}}} choices for the vowel to include, and
{{{5}}} positions to place the chosen vowel,
there are {{{5*5*4*3*2=600}}} possible words.
Calculating it a different way,
You could make {{{5}}} different sets of 5 letters including all consonants and 1 vowel.
For each of those sets,
there are {{{5!=5*4*3*2=120}}} ways to arrange them,
making {{{5*5!=5*120=600}}} 5-letter words.
 
(iii) If those two letters are preselected for first and last letter,
there are {{{9-2=7}}} lettered to choose the 3 middle letters from.
That gives us {{{7*6*5=highlight(210)}}} choices.
 
(iv)  There are {{{4}}} choirs of consonant for the first letter.
There are {{{7!=7*6*5*4=840}}} choices for the sequence of 4 different letters to follow.
So, there are {{{4*840=3360}}} 5-letter words
starting with s consonant, with no repeated letters.
 
(v) Using all 9 letters,
There are {9!=9*8*7*6*5}}} 5-letter words that can need formed with no repeated letters.
Excluding S, you could only form
{8!=8*7*6*5*4}}} .
So, there are
{{{9*8*7*6*5-8*7*6*5*4=8*7*6*5*(9-4)=8*7*6*5*5=840}}}
words containing S.
 
(vi) If you want vowels and consonants to alternate,
you can start with a vowel or a consonant.
If you start with a vowel, you will use 3 vowels,
in one of {{{5*4*3=60}}}
different arrangements.
You will have {{{4*3=12}}} possible arrangements of 2 different consonants to fill the 2nd and 4th letter positions.
That makes {{{60*12=720}}} different words alternating voweks and consonants, staring with a vowel.
Starting with a consonant,
you would have {{{4*3*2=24}}} different arrangements of 3 consonants,
and {{{5*4=20}}} different arrangements of 3 vowels,
For a total of {{{24*20=480}}} 5-letter words,
alternating vowels and consonants starting with a consonant.
In all, you would have {{{720+480=1200}}} 5-letter words alternating vowels and consonants.