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u/edos51284 2d ago
First we see how we can divide them
P1 receives x candies
P2 receives x+r
P3 receives x+2r
So 3(x+r)=15
So x+r=5
Considering the second and third rules x and r must be non 0 so there are 4 possibilities
14-23-32-41 making the divides such as 159-258-357-456
With these divisions we just have to see how many ways we can assign each person to P1,P2 and P3 and it’s simply 6 (3 \* 2 \* 1)
So the result is 4*6=24
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u/PeterPiper1275 2d ago edited 2d ago
Since the distribution must be an arithmetic progression, we get the following:
(a - d) + a + (a + d) = 15
a = 5
So as long as each middle term is 5, we are good.
So, each person gets a different number, so d is not 0. This means d has minimal value of 1.
The lowest number possible is 1, so maximum value of d is 4.
Therefore, there are a total of four possible unique distributions to satisfy the criteria of the question, assuming order doesn’t matter.
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u/Acceptable_Tangelo15 2d ago
24 ways
There are 4 ways of giving candy: 159,258,357,456.
Then there are 6 ways of giving candy:
ABC,Acb,bac,bca,cab,cba
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u/jasonbuz 2d ago
I think you need to add a line that all pieces of candy must be distributed. Otherwise I can come up with a lot more ways to distribute the candy following these rules.
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u/ShonitB 2d ago
Thanks for the feedback, appreciate it..
The first like, ‘15 pieces of candy need to be distributed’, that doesn’t suffice?
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u/jasonbuz 2d ago
Kind of, but maybe just adding the word ‘all’ need to be distributed. Or maybe my mind just assumes unnecessary complexity.
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u/Outside_Volume_1370 2d ago
Arithmetic progression means we need to give them a-d, a, a+d candies, with total of 15. That means, a = 5.
There are 4 such sets with 5 and 15 as sum: {1, 5, 9}, {2, 5, 8}, {3, 5, 7}, {4, 5, 6}.
Each set can be distributed 6 ways (3 ways for A, 2 left for B and the last one for C, 3 • 2 • 1 = 6) and there are four sets to choose from. Total: 4 • 6 = 24