Question 1160355: Determine the number of ways of placing the numbers 1-9 in a circle, so that the sum of any three numbers in consecutive positions is divisible by 3 (Two arrangements are considered the same if one arrangement can be rotated to obtain the other.)
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Answer by greenestamps(13206)   (Show Source):
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The nine numbers can be divided into three groups:
A: 1, 4, 7 (1 more than a multiple of 3)
B: 2, 5, 8 (1 less than a multiple of 3)
C: 3, 6, 9 (a multiple of 3)
A sequence of three numbers, one chosen from each group, will have a sum that is divisible by 3. Since there are the same number of numbers in each group, the arrangement of the numbers around the circle must be ABCABCABC.
We can count the number of different arrangements by counting the numbers of ways we can choose the number for each position, going around the table one place at a time.
(1) We can choose any of the 9 numbers first. 9 choices.
(2) The second number must be one of the 6 numbers in the two other groups. 6 choices.
(3) The third number must be one of the 3 numbers in the third group. 3 choices.
(4) The fourth number must be one of the remaining 2 numbers in the first group. 2 choices.
(5) The fifth number must be one of the remaining 2 numbers in the second group. 2 choices.
(6) The sixth number must be one of the remaining 2 numbers in the third group. 2 choices.
(7) The seventh, eighth, and ninth numbers must be the 1 remaining numbers in the first, second, and third groups, respectively. 1 choice each.
The total number of arrangements is the product of all the numbers of choices:
9*6*3*2*2*2*1*1*1 = 1296
However, two arrangements which are the same except for a rotation are considered to be the same. That essentially means we don't know where the "starting point" is for the arrangement; and that means our count of 1296 is too large by a factor of 9.
So with the given rules, the number of arrangements is 1296/9 = 144.
ANSWER: 144 different arrangements
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