Emma's Dilemma
Aim:
The aim of this investigation is to study the number of combinations, which can be used for different words. I will examine words where the letters are all different or where there may be more than one repeated letter. I will examine words with different numbers of letters. I will then eventually work out a formula for it.
Investigation:
In the name LUCY there are four different letters and there are 24 different arrangements:
.
. LUCY
2. LUYC
3. LCUY
4. LCYU
5. LYCU
6. LYUC
7. ULYC
8. ULCY
9. UCLY
0. UCYL
1. UYCL
2. UYLC
3. CLUY
4. CLYU
5. CYUL
6. CYLU
7. CUYL
8. CULY
9. YLUC
20. YLCU
21. YCLU
22. YCUL
23. YUCL
24. YULC
In the three-letter word SAM, there are 3 different letters and 6 different arrangements:
.
. SAM
2. SMA
3. MAS
4. MSA
5. ASM
6. AMS
In the two-letter word, MO there are 2 different letters and 2 different arrangements:
.
. MO
2. OM
Number of all different Letters
Number of Different Arrangements
2
2
3
6
4
24
5
20
6
720
7
5040
If we put all of this data into a table;
From the above table of results, I have found out that a two-letter word has two arrangements, and a three-letter word has six arrangements. If we take, for example, a three-letter word, I have worked out that if we do 3 (which is the length of the word) ? 2 ?1= 6 which is the number of different arrangements. In a four letter word, to work out the amount of different arrangements, you can do 4 ? 3 ? 2 ? 1= 24. As you can see from the above table that this is correct. You can also do this process on a calculator, by using the factorial button, which looks like ! So four factorial (4!) is the same as 4 ? 3 ? 2 ? 1. So by using factorial (!) I can predict that there will be 40320 different arrangements for an eight-letter word. If n= the number of letters in the and a= the number of different arrangements then the formula that I worked out is n!
Aim:
The aim of this investigation is to study the number of combinations, which can be used for different words. I will examine words where the letters are all different or where there may be more than one repeated letter. I will examine words with different numbers of letters. I will then eventually work out a formula for it.
Investigation:
In the name LUCY there are four different letters and there are 24 different arrangements:
.
. LUCY
2. LUYC
3. LCUY
4. LCYU
5. LYCU
6. LYUC
7. ULYC
8. ULCY
9. UCLY
0. UCYL
1. UYCL
2. UYLC
3. CLUY
4. CLYU
5. CYUL
6. CYLU
7. CUYL
8. CULY
9. YLUC
20. YLCU
21. YCLU
22. YCUL
23. YUCL
24. YULC
In the three-letter word SAM, there are 3 different letters and 6 different arrangements:
.
. SAM
2. SMA
3. MAS
4. MSA
5. ASM
6. AMS
In the two-letter word, MO there are 2 different letters and 2 different arrangements:
.
. MO
2. OM
Number of all different Letters
Number of Different Arrangements
2
2
3
6
4
24
5
20
6
720
7
5040
If we put all of this data into a table;
From the above table of results, I have found out that a two-letter word has two arrangements, and a three-letter word has six arrangements. If we take, for example, a three-letter word, I have worked out that if we do 3 (which is the length of the word) ? 2 ?1= 6 which is the number of different arrangements. In a four letter word, to work out the amount of different arrangements, you can do 4 ? 3 ? 2 ? 1= 24. As you can see from the above table that this is correct. You can also do this process on a calculator, by using the factorial button, which looks like ! So four factorial (4!) is the same as 4 ? 3 ? 2 ? 1. So by using factorial (!) I can predict that there will be 40320 different arrangements for an eight-letter word. If n= the number of letters in the and a= the number of different arrangements then the formula that I worked out is n!
