Counting problem involving anagrams

In summary, the problem asks how many anagrams can be made with 4 distinct letters, two of which must be a, b, or c, using the first ten letters of the alphabet. The solution involves selecting two letters from the three options of a, b, and c, and then selecting two more letters from the remaining seven options. These four letters can then be arranged in 4! ways, giving a total of 1512 possible anagrams.
  • #1
V0ODO0CH1LD
278
0

Homework Statement



How many anagrams with 4 distinct letters and that have two of the letters "a", "b" and "c" can you make using the first ten letters of the alphabet?

Homework Equations





The Attempt at a Solution



First I assume that by anagram they mean letters arranged in any order (like as if any two lettered word had two anagrams and any three lettered word had six).

Now as far as solving the problem goes, I don't even know where to start.. I though of using factorials for the 4 distinct letters, but I'm not even sure of that.
 
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  • #2
V0ODO0CH1LD said:

Homework Statement



How many anagrams with 4 distinct letters and that have two of the letters "a", "b" and "c" can you make using the first ten letters of the alphabet?

Homework Equations



The Attempt at a Solution



First I assume that by anagram they mean letters arranged in any order (like as if any two lettered word had two anagrams and any three lettered word had six).

Now as far as solving the problem goes, I don't even know where to start.. I though of using factorials for the 4 distinct letters, but I'm not even sure of that.
An anagram of four letters is simply a four character string of letters of the alphabet. It doesn't necessarily spell anything.
 
  • #3
V0ODO0CH1LD said:

Homework Statement



How many anagrams with 4 distinct letters and that have two of the letters "a", "b" and "c" can you make using the first ten letters of the alphabet?

Homework Equations





The Attempt at a Solution



First I assume that by anagram they mean letters arranged in any order (like as if any two lettered word had two anagrams and any three lettered word had six).

Now as far as solving the problem goes, I don't even know where to start.. I though of using factorials for the 4 distinct letters, but I'm not even sure of that.

Think about how many ways you can pick two letters from a,b,c and how many ways to pick the other two and how many ways to arrange them.
 
  • #4
I thought about starting from simpler scenarios and work from there. Like; if I just wanted all the anagrams that can be formed using the first ten letters of the alphabet it would be 10^10; right? But then I want to subtract all the anagrams that have 7, 8, 9 or 10 repeated letters.

For the 10 repeated letters it's easy, just one anagram for each letter that's 10.
aaaaaaaaaa
bbbbbbbbbb
...
jjjjjjjjjjjjjjjjjjj

For 9 it's 10 for each letter, but for each 10 of those there is 9 options for the 10th unrepeated letter. So..

aaaaaaaaab
aaaaaaaaba
...
baaaaaaaaa
...
aaaaaaaaaj
aaaaaaaaja
...
jaaaaaaaaa

So far I have 10 + 900. But I think I am overcomplicating things. There must be an easier way to do this..
 
  • #5
V0ODO0CH1LD said:
I thought about starting from simpler scenarios and work from there. Like; if I just wanted all the anagrams that can be formed using the first ten letters of the alphabet it would be 10^10; right? But then I want to subtract all the anagrams that have 7, 8, 9 or 10 repeated letters.

For the 10 repeated letters it's easy, just one anagram for each letter that's 10.
aaaaaaaaaa
bbbbbbbbbb
...
jjjjjjjjjjjjjjjjjjj

For 9 it's 10 for each letter, but for each 10 of those there is 9 options for the 10th unrepeated letter. So..

aaaaaaaaab
aaaaaaaaba
...
baaaaaaaaa
...
aaaaaaaaaj
aaaaaaaaja
...
jaaaaaaaaa

So far I have 10 + 900. But I think I am overcomplicating things. There must be an easier way to do this..
The problem states that the anagrams have 4 letters, and the letters are distinct. -- No repeats and only 4 letters in length.
 
  • #6
Ah! Damn my interpretation skills.

Okay; so I have three ways to select two letters out three. Then I have 12 ways to organize those in a four character string (taking into consideration order matters). That gives me 36 ways to get two of the letters a, b and c in my anagram; right? But I still have 2 spots left and for the first I could have any of the eight remaining letters minus the one left out from a, b and c. And finally for the last spot I have six options.

So that is (3*12)*7*6 = 1512. Is that correct or did I forget something?
 
  • #7
V0ODO0CH1LD said:
Ah! Damn my interpretation skills.

Okay; so I have three ways to select two letters out three. Then I have 12 ways to organize those in a four character string (taking into consideration order matters). That gives me 36 ways to get two of the letters a, b and c in my anagram; right? But I still have 2 spots left and for the first I could have any of the eight remaining letters minus the one left out from a, b and c. And finally for the last spot I have six options.

So that is (3*12)*7*6 = 1512. Is that correct or did I forget something?

That is correct. But it is more direct expressed like this:$$
\binom 3 2 \binom 7 2 4!$$where the first binomial is the number of ways of choosing the 2 from the 3, the next binomial is the number of ways of choosing the other 2 from the remaining 7, and the factorial is the number of ways to arrange the 4 after you have them.
 
  • #8
Okay! Thanks!
 

FAQ: Counting problem involving anagrams

1. What is an anagram?

An anagram is a word or phrase that is formed by rearranging the letters of another word or phrase. For example, "listen" and "silent" are anagrams of each other.

2. How do you approach a counting problem involving anagrams?

The first step is to identify the distinct letters in the given word or phrase. Then, determine the number of times each letter appears and use this to calculate the total number of possible arrangements. Finally, consider any duplicate letters and divide the total number of arrangements by the factorial of the number of duplicates to account for repeated arrangements.

3. Can the same letter be used more than once in an anagram?

Yes, the same letter can be used multiple times in an anagram. This is why it is important to consider the number of times each letter appears when calculating the total number of possible arrangements.

4. Are there any shortcuts or formulas for solving anagram counting problems?

Yes, there are certain formulas that can be used to solve anagram counting problems more efficiently, such as the multinomial theorem. However, these may only apply in specific cases and may not be applicable to all anagram counting problems.

5. Can computers be used to solve anagram counting problems?

Yes, computers can be programmed to solve anagram counting problems by using algorithms and data structures. This can be especially useful for larger or more complex anagrams that may be difficult to solve manually.

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