GCSE Mathematics: Emma's Dilemma
GCSE Mathematics: Emma's Dilemma
Emma and Lucy are playing with arrangements of their names. One arrangement of LUCY is L.U.C.Y. Another arrangement is Y.L.C.U. We are investigating the number of different arrangements of Emma and Lucy's names. Then we shall move onto the number of different arrangements of various groups of letters.
Firstly, I shall list all the possible combinations for the word "LUCY":
LUCY LCUY LYUC LYCU LUYC LCYU UCYL UYCL ULCY ULYC
UYLC UCLY CYLU CYUL CULY CUYL CLYU CLUY YCUL YCLU
YUCL YULC YLUC YLCU
= 24 Combinations
I listed all these combinations by using the following method:
Step 1
Arrange the listing process into 4 stages:
1 _ _ _
2 _ _ _
3 _ _ _
4 _ _ _
Step 2
You start off with the original word: 1234
Acquire the combinations of the last two numbers first and you end up with 1243.
Now that you have 1243, do the last three numbers and try the different possibilities:
423 1432 1342 1324.
Because the number 2 has been the first number of last three numbers, we don't do it again.
Step 3
Now that we have listed all the arrangements beginning with 1, we do the list of arrangements with 2 in the beginning:
Start off with 2134 and do same thing to it, it will look like this:
2134 2143 2431 2413 2314 2341
Step 4
Now that we have established the different arrangements of 2, we go ahead and do 3 at the beginning:
3124 3142 3241 3214 3412 3421
Step 5
Now that we have established the different arrangements of 3, we go ahead and do 4 at the beginning:
4123 4132 4231 4213 4312 4321
By using the above method you will definitely acquire all the different arrangements of a four letter word with different letters.
Now I shall list all the possible combinations for the word "EMMA". But there is a problem! There are 2 of the same letter (M) in "EMMA". To start off with, I shall use the same method that I used with the word "LUCY".
EMMA EMAM EMMA WMAM EAMM EAMM MMAE MAME MEMA
MEAM MAEM MMEA MAEM MAME MMEA MMAE MEAM MEMA
AMME AMEM AMME AMEM AEMM AEMM
=24 Combinations
Obviously seeing as "LUCY" & "EMMA" are both 4 letter words; by using the same process I acquired the same amount of results (24). There is however, another method. I could treat both the M's as one collective term. Therefore if you swapped them around, you would get the same word or a repeat of a previous combination, so you cannot include it in your list of combinations. So by treating both the M's as 1 collective term I would NOT find 24 combinations. How many would I find?
I shall list all the possible combinations of "EMMA" using this method:
EMMA EMAM EAMM MMEA MMAE MAEM MAME AMEM AMME AEMM MEMA MEAM
= 12 Combinations
I acquired 12 combinations by using this method. This is because in the previous case when we found 24 combinations, each M was treated as if it was a different letter, therefore allowing us to find more arrangements. Whereas in this method we could not swap the M's around as we could do before. Now that we have investigated 4 letter words, let's move onto 5 letter words:
I am going to number each letter and list the combinations as numbers. The 1st letter of the word will be numbered as 1, 2nd as 2, 3rd letter as 3 etc.
2345 13245
2354 13254
2435 --- 6 arrangements 13452 --- 6 arrangements
2453 13425
2534 13524
2543 13542
4253 15423
4352 --- 6 arrangements 15324 ---- 6 arrangements
4523 15342
4532 15243
4325 15234
Have you noticed that the combinations of last 4 numbers added up equal 24, so if we multiply 24 by 5, and get 120, then 120 should be the total of arrangements.
Carry on and investigate if a number has 6 figures, then the total of different combinations should be 120 times by 6, and get 720, and 720 should be the total of arrangements.
Carry on, if a number has 7 figures, then the total of different combinations should be 720 times by 7, and get 5040, the total of combinations should be 5040.
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Have you noticed that the combinations of last 4 numbers added up equal 24, so if we multiply 24 by 5, and get 120, then 120 should be the total of arrangements.
Carry on and investigate if a number has 6 figures, then the total of different combinations should be 120 times by 6, and get 720, and 720 should be the total of arrangements.
Carry on, if a number has 7 figures, then the total of different combinations should be 720 times by 7, and get 5040, the total of combinations should be 5040.
This is my prediction, let's investigate and try to extract a formula to confirm my prediction:
Number Of Letters Number Of Combinations
3 6
4 4 x 6
5 4 x 6 x 5
6 4 x 6 x 5 x 6
7 4 x 6 x 5 x 6 x 7
8 4 x 6 x 5 x 6 x 7 x 8
We can re-write this as:
Number Of Letters Number Of Combinations
2 2 x 1
3 3 x 2 x 1
4 4 x 3 x 2 x 1
5 5 x 4 x 3 x 2 x 1
6 6 x 5 x 4 x 3 x 2 x 1
7 7 x 6 x 5 x 4 x 3 x 2 x 1
8 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
Eureka! There is a mathematical operation which does exactly what I have been doing in the above tables. The name of it is "Factorial" and its symbol is (!), which you can find on your calculator.
Summary
When you multiply the previous number of combinations by the amount of letters in your new word, you can predict the next number of combinations. Why?
Explanation
Start off with "LUCY". You have 4 blank spaces:
_ _ _ _ Put "L" into one space
L _ _ _ You now only have 3 spaces to fit "U" into
L U _ _ Now you only have 2 spaces for C
L U C _ So obviously you now only have 1 space left for Y.
Therefore the more letters you put in, the less options you have for the remaining letters. So you start off with 4 spaces, then 3 spaces, then 2 spaces, then 1 space.
4 x 3 x 2 x 1 = 4! (4 Factorial) = 24 (The number of combinations for a 4 letter word)
Let "n" be the number of letters in a word. The formula is then n! which can be calculated using your calculator by pressing the number which is represented by "n" and ! (n!)
Let's confirm this formula, if a word has
letter: It has 1 arrangement formula: 1 x 1 = 1 It works
2 letters: It has 2 arrangements formula: 1 x 2 = 2 It works
3 letters: It has 6 arrangements formula: 1 x 2 x 3 = 6 It works
4 letters: It has 24 arrangements formula: 1 x 2 x 3 x 4 = 24 It works
My Formula is now confirmed to be correct = n!
For a four letter word with duplicated letters, we have to construct a formula. Lets conduct the same investigation:
Use For example: 223, 334
Let's try to work out the combinations using the same method as before:
223 can be arranged as 232, 322
There are only 3 arrangements.
Try 4 figures with 2 of the same number.
223
which can be arranged as:
223 2123 3122 1232 2132 3212 1322 2213 3221 2231 2312 2321
The total of arrangements is 12.
Try 5 letters:
42213 12234
42231 12243
42123 12324
42132 12342
42321 12423
42312 12432
41223 13224
41232 13242
41322 13422
43122 14223
43212 14232
43221 14322
21234 23124
21243 23142
21324 23214
21342 23241
21423 23421
21421 23412
22134 24123
22143 24132 32412
22314 24231 32421
22341 24213 34122
22413 24312 34212
22431 24321 34221
32241
32214
32142
32124
31422
31242
31224
So the total arrangements are 12 x 5=60
Let's put our results in a table and try to extract a formula from there:
Number Of Letters (With 2 of the letters the same) Number Of Combinations
2 1
3 6
4 12
5 60
Which can be re-written as:
Number Of Letters (With 2 letters the same) Number Of Combinations
2 1
3 3 x 1
4 4 x 3 x 1
5 5 x 4 x 3 x 1
Let's work out the formula:
if n= number of figures
a= number of arrangements
the formula is a=n!/2
Let's confirm the formula:
Number Of Letters (with 2 letters the same) Number Of Combinations
2 2/2=1 It works
3 (1 x 2 x 3) / 2=3 It works
4 (1 x 2 x 3 x 4) / 2=12 It works
Now I shall investigate different amounts of letters with different amounts of repeated letters. I shall start off with a four letter word where 3 of the letters are the same. For this I shall use the combination "3331":
3331 3313 3133 1333 - There are 4 combinations.
Now let's try a 5 letter word with 3 letters the same:
33312 31233
33321 31323
33123 31332
33132 32331
33231 32313
33213 32133
21333 12333
23133 13233
23313 13323
23331 13332 = 20 Combinations
Let's try 6 letters with 3 of the same letter:
333124 332134
333142 332143
333214 334321
333241 332314
333412 332341
333421 332413
331234 332431
331243 334123
331324 334132
331342 334213
331423 334231
331432 334312
312334 321334
312343 312433
313234 313243
313324 313342
313423 313432
314233 314323
23334 133324
23343 133342
23433 133234
24333 133243
24332 133423
34323 133432
34233 142333
32334 143233
32343 143323
32433 143332
314332
The total of arrangements for 6 letters with 3 of the same letter is 120, 20 x 6
Let's put this in a table:
Number Of Letters (3 of the same letter) Number Of Combinations
3 1
4 1*4
5 1*4*5
6 1*4*5*6
If a = number of arrangements
n = number of letters
The formula for a word with 3 of the same letters is a= ni/6
Let's review the formulae:
Formula for a word with different letters:
a=n!
Formula for a word with 2 of the same letter:
a=n!/2
Formula for a word with 3 of the same letter:
a=n!/6
I shall put them in a table for easier analysis:
n
2
3
x
2
6
n represent the number of figures of a number
x represent the number which n! is being divided by
Do you notice that x equal the last x multiplied by n?
Using this information, I predict that the formula for a word with 4 of the same letter is:
a= n!/24
Let's confirm it:
Try a four letter word with 4 of the same letter:
a=n/24=(1 x 2 x 3 x 4)/24=1 The formula works
Try a 5 letter word with 4 of the same letter:
1112
1121
1211 ----- 5 arrangements
2111
21111
a=n/24=(1 x 2 x 3 x 4 x 5)/24=5 The formula works
This confirms my formula!
Let's investigate the formula, and improve it.
n
2
3
4
5
x
2
6
24
10
Looking at the table, do you notice how X = the factorial of N? Using this information, we can construct a formula.
Let "a" represent the number of arrangements
"n" represent the number of figures
"x" represent the number of duplicated letters
The formula will be a=n!/x! (! Cannot be cancelled out)
Now I shall investigate what would happen if a word has 2 pairs of the same letter:
Let's use a 4 letter word.
122 2122
212 2212
221 2221
Let's see if the formula still works
a=(1 x 2 x 3 x 4)/2=12
No, it doesn't work unless we divide it by 2.
Let's try 6 letters, with 2 pairs of 3 of the same number:
11222 121212 222111 211212
12122 121122 221211 211221
12212 122112 --10 arrangements 221121 212112 ---10 arrangements
12221 122121 221112 212121
21221 122211 211122 212211
Total of arrangements is 20
Let's look at the formula:
a=(1x2x3x4x5x6)/1x2x3 =120 NO, it doesn't work but unless we divide it by another 6 which is (1 x 2 x 3)
Explanation
There are a different number of extra arrangements caused by each repeated letter for example if a name has 2 As and 3 Bs. So we first divide n! by x1! to get rid of all the extra combinations caused by the first repeated letter. Then we divide it by x2! To get rid of all the extra combinations caused by the 2nd repeated letter, and so on. So if you divide the factorial of the number of repeated letters in the word by the number of letters, the formula will work.
For example:
Using a four letter word with 2 pairs of 2 of the same letter:
a= (1 x 2 x 3 x 4)/(1 x 2 x 1 x 2)= 6 it works
So I predict it would still work for 6 letters with 2 pairs of 3 of the same number:
Following this formula, I predict the arrangement for this is:
a=(1 x 2 x 3 x 4 x 5 x 6)/1 x 2 x 3 x 1 x 2 x 3=20
11222 121221 122121
12122 121212 122211
12212 121122
12221 122112
Total Of Arrangements= 20
The formula works.
I predict the arrangements for 8 letters will be a= (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8)/1 x 2 x 3 x 4 x 1 x 2 x 3 x 4=70
Let's confirm it:
1112222 11211222 11221212
1121222 11212122 11221221
1122122 11212212 11222112
1122212 11212221 11222121
1122221 11221122 11222211
2111222 12122112 12212121 12121212
2112122 12122121 12212211 12211212
2112212 12122211 12221112
2112221 12211122 12221121
2121122 12211221 12221211
2121221 12212112 12222111
Total Of arrangements is 70, it works
The formula is confirmed
This formula can be written as:
a=n!/x!x!
x represents the number of same letter
What about a word with a different number of same letters.
For example:
1122, 111122
Let's see if the formula still works.
The formula is a=n!/x!x!
We need to change the formula, because there are 2 pairs of same numbers with different number of figures. So we change the formula to a=ni/x1i * x2i
Let's try 5 figures with 3 same of the same letter, and 2 same number.
According to the formula, I predict the total arrangement for this is
a=(1 x 2 x 3 x 4 x 5)/(3 x 2 x 1 x 2 x 1)=10
1122 12211
1212 21112
1221 21121 12112 21211
2121 22111
The formula still works.
Let's try 7 letters, with 3 of the same letter and 4 of the same letter
I predict that the total of arrangements is:
a=(1 x 2 x 3 x 4 x 5 x 6 x 7)/(3 x 2 x 1 x 4 x 3 x 2 x 1)=35
112222 1222211
2222111 2211212
121222 1222121
2221211 2211221
122122 1222112
2221121 2212112
122212 1221221
2221112 2212121
122221 1221212
2211122 2212211
212221 1221122
212212 1212122
211222
2111222 2112221 2121221 2122211
2112122 2121122 2122112
2112212 2121212 2122121
Total of arrangements is 35, the formula works.
The formula is confirmed.
Conclusion
From the investigation above we find out the formula for calculating the number of arrangements, it's
a=n!/x!
a represent the total arrangements
n represent the number of figures of the number
! represents the factorial function.
x represent the number of duplicated letters in the word
for 2 pairs of the same letter:
the formula is a=n!/x!x!
for 2 pairs of different duplicated letters of same number in a word:
the formula is a=n!/x1!x2!
for 3 pairs of same number of duplicated letters in a word:
the formula is a=n!/x!x!x!
for 3 pairs of a different number of the same letter in a word:
The formula is a=n!/x1!x2!x3!.
The formula can be also used to find the amount of different arrangements of letters.
For example:
xxyy
the arrangement for this is a=(4 x 3 x 2 x 1)/(2 x 1 x 2 x 1)=6
xxyyy
the arrangement for this is a=(5 x 4 x 3 x 2 x 1)/(3 x 2 x 1 x 2 x 1)=10
xxxxxxyyyyyyyyyy
The formula for this is:
a=n!/x1!x2!=(16 x 15 x 14 x 13 x 12 x 11 x 10 x 9x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 ) / ( 1 0 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 3 x 2 x 1 x 6 x 5 x 4 x 3 x 2 x 1 ) = 8 0 0 8
The total arrangement is 8008.
Using this formula, we can find out the total arrangements of all numbers and letters.
Sheroy Zaq (X2) Mrs Whybrow