How many different combinations of 3 letters which can be A, B, or C are there, if the ordering doesn't matter? The total number if the order does matter is 27=3*3*3. The number of groups where the order doesn't matter is 10. I wrote them out just to make sure: AAA, AAB, AAC, ....etc. I eliminated the repeated groups such as with ABB, BAB, BBA, etc. What is the formula to arrive at 10? I can't figure it out.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Should be easy problem
- Thread starter ryan124
- Start date
How many different combinations of 3 letters which can be A, B, or C are there
What you are asking about are Permutations (order matters) and Combinations (order does not matter). Here is one site that explains in good detail, with examples:
http://www.mathsisfun.com/combinatorics ... tions.html
Hello, ryan124!
I have a very primitive approach.
There are three types of combinations:
. . \(\displaystyle \begin{array}{ccc}(1) & XXX & \text{Triple} \\ (2) & XXY & \text{Pair} \\ (3) & XYZ & \text{One-of-each} \end{array}\)
\(\displaystyle (1)\;XXX\)
\(\displaystyle \text{We must choose a letter for the }X.\)
\(\displaystyle \text{There are: }3\text{ Triples.}\)
\(\displaystyle (2)\;XXY\)
\(\displaystyle \text{There are 3 choices for the }X\text{ and 2 choices for the }Y.\)
\(\displaystyle \text{Hence, there are: }3\cdot2\:=\:6\text{ Pairs.}\)
\(\displaystyle (3)\;XYZ\)
\(\displaystyle \text{We must choose one of each letter.}\)
\(\displaystyle \text{There is: }\;1\text{ One-of-each.}\)
\(\displaystyle \text{Therefore, there are: }\:3 + 6 + 1 \:=\:10 \text{ combinations.}\)
How many different combinations of 3 letters which can be A, B, or C are there,
if the ordering does not matter?
The total number if the order does matter is: .\(\displaystyle 3\cdot3\cdot3 \:=\:27\)
The number of groups where the order does not matter is 10.
I wrote them out just to make sure: AAA, AAB, AAC, ....etc.
I eliminated the repeated groups such as with ABB, BAB, BBA, etc.
What is the formula to arrive at 10?
I have a very primitive approach.
There are three types of combinations:
. . \(\displaystyle \begin{array}{ccc}(1) & XXX & \text{Triple} \\ (2) & XXY & \text{Pair} \\ (3) & XYZ & \text{One-of-each} \end{array}\)
\(\displaystyle (1)\;XXX\)
\(\displaystyle \text{We must choose a letter for the }X.\)
\(\displaystyle \text{There are: }3\text{ Triples.}\)
\(\displaystyle (2)\;XXY\)
\(\displaystyle \text{There are 3 choices for the }X\text{ and 2 choices for the }Y.\)
\(\displaystyle \text{Hence, there are: }3\cdot2\:=\:6\text{ Pairs.}\)
\(\displaystyle (3)\;XYZ\)
\(\displaystyle \text{We must choose one of each letter.}\)
\(\displaystyle \text{There is: }\;1\text{ One-of-each.}\)
\(\displaystyle \text{Therefore, there are: }\:3 + 6 + 1 \:=\:10 \text{ combinations.}\)
I'm too lazy on a Sunday!
Go here: http://www.research.att.com/~njas/sequences/index.html
Enter A001700 in search box
Go here: http://www.research.att.com/~njas/sequences/index.html
Enter A001700 in search box