EMMA’s Dilemma - page 4
Keywords: EMMA’s Dilemma
By Grant867 on 07/12/2008
Level: GCSE Key Stage 4 (Years 10-11)
Page Number: 4 of 5 pages: 1 2 3 4 5
figures. so we change the formula to a=ni/x1i*x2i
Let’s try 5 figures with 3 same number, and 2 same number.
According to the formula, I expect the total arrangement for this is
a=(1*2*3*4*5)/(3*2*1*2*1)=10
11122 12211
11212 21112
11221 21121 -------10 arrangements
12112 21211
12121 22111
The formular still works.
Let’s try 7 fig, with 3 same number, and 4 same number.
I expect the total arrangement is
a=(1*2*3*4*5*6*7)/(3*2*1*4*3*2*1)=35
1112222 1222211 2222111 2211212
1121222 1222121 2221211 2211221
1122122 1222112 2221121 2212112-----10 arrangements
1122212 1221221-----15 arrangements 2221112 2212121
1122221 1221212 2211122 2212211
1212221 1221122
1212212
1212122
1211222
2111222 2112221 2121221 2122211
2112122 2121122 2122112 --------- 10 arrangements
2112212 2121212 2122121
Total arrangment is 35, the formular works.
The formular is confirmed.
What about three pairs of same number
The formmular need to be rewritten as a=ni/xixixi
There are three xi need to mutiple ni, because there are three pairs of same number.
if there are two pair of same number of figures of same number, then there are only two xi need to mutiply, and if there are two pair of different number of figures of same number, then there would be x1i and x2i need to mutiply.
Let’s confirm the formular.
let’s try 112233, according the formula, I expect the total arrangements is a=(1*2*3*4*5*6)/(1*2*1*2*1*2)=90
Let’s confirmed
112233 121233 123123 131223 132231
112323 121323 123132 131232 132123
112332 121332 123213 131322 132132 ------ 30 arrangements
113223 122133 123231 132321 133122
113232 122313 123312 132312 133212
113322 122331 123321 132213 133221
2------- 3------
so on ------30 arrangements so on --------- 30arrangements
The total arrangement is 90, the formular works.
Formular is confirmed
What about three pairs of different number of figures of a number
For example:
122333
according the formula, the total arrangment is
a=(1*2*3*4*5*6)/(1*1*2*1*2*3)=60
Let’s confirm it:
122333 212333 231332 3--------
123233 213233 232133 so on ------- 30 arrangements
123323 213323 232313
123332 213332 232331 -----30 arrangements
132233 221333 233123
132323 223133 233132
132332 223313 133213
133223 223331 233231
133232 231233 233312
133322 231323 233321
The formular works
Formular is confirmed
From the investigation above we find out the formular for calculating the number of arrangements, it’s
a=ni/xi
a represent the total arrangements
n represent the number of figures of the number
I represent the key I
x represent the numbers of figures of same number of the number
if there are more than one pair of same number, x2, or x3, so on may added to the formular, it depend how many pairs of same number.
For example:
for 2 pairs of same number of figures of same number of a number
the formula is a=ni/xixi
for 2 pairs of different number of figures of same number of a number
the formula is a=ni/x1ix2i
for 3 pairs of same number of figures of same number
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