As children, we all learned to count. Thus we rarely think about counting as ‘difficult’. Yet mathematicians have developed a special branch of mathematics for the art of counting. The branch is called Combinatorics. Typical questions in Combinatorics are:
1) In how many ways can a stack of 52 playing cards be arranged?
2) When we have a vase with 5 black and 5 white balls, in how many sequences can we pull them out?
In today’s problems, we work with the cards 1 to 9:
1) How many ways?*/*****
It is easy to arrange these cards into 3 groups, all with the same sum:
One reason it is easy, is because there are several solutions. Withe sum of the 9 cards being 45, each of the three groups will have to have sum 15. But in how many ways exactly can we divide the cards 1-9 into three groups, all with the same sum?
2) Be creative****/*****
Now be creative in the arrangement of your cards. In how many ways can you create 3 groups in such a way that the three groups still all have the same sum, but the sum is not 15?
Yeah, you may cheat in this problem. But your cheating is limited to arranging the cards.
3) Combinatorics (unsolved)*****/*****
The sum of the first n cards is n(n+1)/2. To divide these number into three groups with the same sum, either n or n+1 mus be a multiple of 3. So this is not possible for n=4, 7, 10 and so on.
Here is a short list
n
2: 0
3: 0
5: 1 (5, 1-4, 2-3)
6: 1 (1-6, 2-5, 3-4)
8: 4
9: see answer to problem 1.
Now can you find a general formula for the number of possible groups?
Or for a simpler start: in how many ways can we draw cards from a series 1-n in such a way that that the sum is some given number?
Or: can you construct an algorithm that shows that the cards 1-3n (n>=2) can always be divided into three groups with the same sum?
I don’t have the answers to these questions, they just look interesting to me.
A new puzzle is published on Fridays, at least twice a month. You may check your solutions here.
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