If the ‘piles’ are unordered then the answer is clearly seven. A simple listing will show that. I do think that is what the problem means.
However, if the ‘piles’ are ordered then the answer is twenty-eight. I don’t think that is what is meant.
Hi,
I have a problem with this exercise - How many ways are there of putting 15 pennies into three piles (each with at least one penny) such that each pile contains an odd number of pennies? I think that the answer is seven, but have difficulties to write it down in a formal way. I would be greateful for any help.
Thanks in advance...
Hello, Gibo!
I agree completely with Plato's explanation.
I tried to find a more formal approach to this problem,
. . but ended up Listing the cases anyway.
Place one penny in each pile: .
Form the other 12 pennies into pairs:.
Now distribute the pairs among the three piles. .There are seven ways:
. .