EMMA’s Dilemma - page 3
Keywords: EMMA’s Dilemma
By Grant867 on 07/12/2008
Level: GCSE Key Stage 4 (Years 10-11)
Page Number: 3 of 5 pages: 1 2 3 4 5
them is this way:
n
1
2
3
x
1
2
6
n represent the number of figures of a number
x represent the divided number in the formular
Do you notice that x equal the last x time n, so I expect the formula for 4 sames number of a number is:
a= ni/24
Let’s confirm it:
try 4 same number.
4 fig, one arrangement.
a=n/24=(1*2*3*4)/24=1 the formular works
try 5 figures
11112
11121
11211 ----- 5 arrangements
12111
21111
a=n/24=(1*2*3*4*5)/24=5 the formular works
So formula is confirmed
Let’s investigate the formula, and improve it.
n
1
2
3
4
5
x
1
2
6
24
110
so the formula for this is x=ni
so if A represent arrangement, and n represent numbers of figures, x represent the number fo same number, and the formula is:
a=ni/xi *notic I can not be cancel out.
What about if a number has 2 pairs of same number. what would happen to the formula.
Let’s try 4 fig with 2 pairs of 2 same number.
1122 2122
1212 2212 ----- 6 arrangements
1221 2221
let’s see if the formula still work
a=(1*2*3*4)/2=12
No, it doesn’t work but if we divide it by two.
let’s try 6 fig, with 2 pairs of 3 same numbers
111222 121212 222111 211212
112122 121122 221211 211221
112212 122112 --10 arrangements 221121 212112 ---10 arrangements
112221 122121 221112 212121
121221 122211 211122 212211
total arrangement is 20
let’s see the formular:
a=(1*2*3*4*5*6)/1*2*3 =120 no, it doesn’t work but if we divide it by another 6 which is (1*2*3)
Can you see the pattern, the formular still work if we times the mutiply again.
For example:
for 4 figures with 2 pairs of 2 same number.
a=(1*2*3*4)/(1*2*1*2)=6 it works
so I expect it still work for 6 figures with 2 pairs of 3 same number.
follow this formula, I predict the arrange for this is a=(1*2*3*4*5*6)/1*2*3*1*2*3=20
111222 121221 122121 2-----
112122 121212 122211 -------- 10 arrangement so on ------10 arrangements
112212 121122
112221 122112
Total arrangement= 20
the formula work
I expect the arrangements for 8 fig will be a= (1*2*3*4*5*6*7*8)/1*2*3*4*1*2*3*4=70
let’s confirm
11112222 11211222 11221212
11121222 11212122 11221221
11122122 11212212 11222112 --------- 15 arrangements
11122212 11212221 11222121
11122221 11221122 11222211
12111222 12122112 12212121 12121212
12112122 12122121 12212211 12211212
12112212 12122211 12221112 --------------20 arrangements
12112221 12211122 12221121
12121122 12211221 12221211
12121221 12212112 12222111
2-------
so on ------35 arrangement
Total arrangement is 70, it works
the formular is confirmed
This formula can be written as:
a=ni/xixi
x represent the number of figures of same number
What about a number with difference number of figure of same number.
For example:
11122,111122
let’s try if the formula still work.
The formula is a=ni/xixi
but we need to change the formula, because there are 2 pairs of same numbers with different number of
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