Joe wants to find all the four-letter words that begin and end with the same letter. How many combinations of letters must he examine? A word is any sequence of letters, such as MTHM, and does not have to be English. (You can use a calculator for this problem.)
the answer is 263, because the first letter and the last letter are the same.
This means that picking a four letter word that follows this rule is just like picking any 3 letter word (pick a 3 letter word, and then add a "copy" of the first letter as a fourth letter)
for example- \(Dye\Rightarrow Dyed\)
so there's a total of 263 =17,576 combinations- 26 options for the first letter, 26 options for the second letter, and 26 options for the third letter.
+675 He would have to look a 264 combinations.
This is because:
2 letter words =
AA to ZZ
AA to AZ is 26 combinations, then you repeat with BA to BZ, CA to CZ etc.
As you repeat this 26 combo thingamajig 26 times,
This means that there are 262 combinations for two letter words.
3 letter words =
AAA to ZZZ
AAA to AAZ is 26 combinations, AAA to AAZ is 26 combinations.
then you repeat with ABA to ABZ, etc.
This means that there are 263 combinations for three letter words.
Therefore..
4 letter words =
AAAA to ZZZZ
AAAA to AAAZ is 26 combinations, then you repeat with AABA to AABZ, AACA to AACZ etc.
This means that there are 264 combinations for four letter words.
Not sure about the answer though...
the answer is 263, because the first letter and the last letter are the same.
This means that picking a four letter word that follows this rule is just like picking any 3 letter word (pick a 3 letter word, and then add a "copy" of the first letter as a fourth letter)
for example- \(Dye\Rightarrow Dyed\)
so there's a total of 263 =17,576 combinations- 26 options for the first letter, 26 options for the second letter, and 26 options for the third letter.
