SOLUTION: We want to build 13 letter “words” using only the first n = 9 letters of the alphabet. For example, if n = 5 we can use the first 5 letters, {a,b, c,d, e} (Recall, words are

Question 1201724: We want to build 13 letter “words” using only the first n = 9
letters of the alphabet. For example, if n = 5 we can use the first
5 letters, {a,b, c,d, e} (Recall, words are just strings of letters,
not necessarily actual English words.)
a. How many of these words are there total?
b. How many of these words contain no repeated letters?
c. How many of these words start with the sub-word “ade”?
d. How many of these words either start with “ade” or end
with “be” or both?
e. How many of the words containing no repeats also do not
contain the sub-word “bed”?

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Answers:

2,541,865,828,329
0
3,486,784,401
34,824,797,289
0

===============================================
Explanation for part (a)

n = 9 letters in the set {a,b,c,d,e,f,g,h,i}

With repeat letters allowed, and 13 slots to work with, we have 9^13 = 2,541,865,828,329 different "words" total.

This massive number is roughly 2.54 trillion.
In scientific notation, it is approximately 2.54 * 10^12

===============================================
Explanation for part (b)

If we had 9 letter words, then a word with no repeated letters could be possible.
Once we reach 10 letters and beyond, we'll have no choice but to repeat a letter.

Therefore, it is impossible to have a 13 letter "word" with no repeated letters if we are picking from the set {a,b,c,d,e,f,g,h,i} that has 9 items.

===============================================
Explanation for part (c)

The word must start with "ade".
We go from 13 slots to 13-3 = 10 slots.

We have 9 letters from the set {a,b,c,d,e,f,g,h,i} and we have 10 slots to fill.

If we can use repeat letters, then we'll have 9^10 = 3,486,784,401 different "words".

===============================================
Explanation for part (d)

If we lock in the last two letters as "be", then we have 13-2 = 11 slots left
9^11 = 31,381,059,609 words that end with "be".

Now let's lock in the first three slots as "ade" and last two as "be".
We have 13-3-2 = 8 slots left
9^8 = 43,046,721

Use these facts:

A = 3,486,784,401 words that start with "ade"
B = 31,381,059,609 words that end with "be"
C = 43,046,721 words that start with "ade" and end with "be"

To be able to say the following
A+B-C
= (3,486,784,401)+(31,381,059,609)-(43,046,721)
= 34,824,797,289

This is the number of 13 letter words where they start with "ade", end with "be", or both.

===============================================
Explanation for part (e)

As mentioned earlier in part (b), we do not have any words where all 13 letters are unique. We'll have repeated letters at some point.

RELATED QUESTIONS
How many 3-letter code words are possible using the first 8 letters of the alphabet if... (answered by scott8148)

How many 3-letter code words are possible using the first 7 letters of the alphabet if... (answered by Alan3354)

Write the word that the number 32 represents. From this word, take 33 out the opposite... (answered by ikleyn)

In mathematics, a word is a list of letters which may or may not form an actual word in... (answered by stanbon)

4. How many 3-letter code words are possible using the first 8 letters of the alphabet if (answered by ramkikk66)

How many seven letter code words can be formed using a standard 26 letter alphabet if... (answered by VFBundy,Edwin McCravy)

In mathematics, a word is a list of letters which may or may not form an actual word in... (answered by Edwin McCravy)

How many ways can you arrange the letters in the word SMILE if the first letter must be... (answered by stanbon)

imagine the alphabet only consisted of 7 letters abcdefg. If each letter can be use once. (answered by Edwin McCravy)