Question

Fool's problem of the day,

A Combinatorial problem,

A square paper sheet is folded four times to get the maximum number of triangles when the paper is unfolded. Find the number of ways of putting \(19 \) identical balls in those triangles which do not contain any other (overlapping) triangle, in such a way that each triangle should contain at least one ball but no triangle contains exactly three balls.

PS:The triangles of the paper are formed by joining the lines formed due the folding of the paper.

Good luck!

Answer 1

1542 ways

Answer 2

Off topic, I'm sad that meta-math was done away :(

Answer 3

@lgbasallote: 1542 ways is not the right answer.

@badreferences: I am sorry to hear that.

Answer 4

oh darn :/ gonna think of a new random number then...

Answer 5

@lgbasallote Problem solving, for the future!

Answer 6

Nevermind. I just needed to read the question more carefully.

Answer 7

Alright, if I folded the paper correctly, we have 16 triangles, and 19 balls. Since each triangle has to have at least one ball, that leaves us with 3 balls left to distribute to 16 triangles such that no triangles gets 2 of those three. Hence, 3 triangles will get 1 ball. That leaves us with an answer of \[\binom{16}{3}=560\]

Answer 8

I'm not entirely positive that 16 triangles is the maximum number of triangles you can get. It's just the most I was able to get.

Answer 9

Forgot one thing. We could distribute all 3 remaining balls to a single triangle, so that means we have \[\binom{16}{3}+\binom{16}{1}=560+16=576\]Ways of doing this.