Curse

Yellow_lid · Oct 02, 2007, 06:11 AM // 06:11

I'm sure there are nerds like me who like to get a bit of a mental workout on the forums...

Here is the hardest question I have encountered in my class so far.

How many ways are there to give three children nine different candies?

DISCLAIMER: Doing math for extended periods of time may cause head pains.

~I DO have the answer, so you aren't doing my HW for me... lol

chowmein69 · Oct 02, 2007, 09:51 PM // 21:51

i think its 729..................

Apoc · Oct 03, 2007, 12:31 AM // 00:31

Quote:

Originally Posted by Soulless Warhawk

i think its 729..................

Correct.
9^3 = 729.

Unless we're going for some probability/statistical maths and using factorials...
Anyway, I don't know the subjects of Math 399 since in Canada, it's probably different/has a different code.

karunpav · Oct 03, 2007, 12:45 AM // 00:45

1, because the kids will eat all of it before you can redistribute it.

Divine Freak · Oct 03, 2007, 01:05 AM // 01:05

Quote:

Originally Posted by karunpav

1, because the kids will eat all of it before you can redistribute it.

LOL

So true

Yellow_lid · Oct 03, 2007, 03:20 AM // 03:20

Right idea, but you didn't quite do it correctly...
Each candy can only go to one child at a time... we shall call them children A, B, and C.
If you only had one candy there would be 3 possibilities (3^1).
A
B
C
If you had two candies how many would there be.
AA
BA
CA
AB
BB
CB
AC
BC
CC
now there are 9 possibilities... notice that I had to to square the number of possibilities when I threw a new candy into the mix (3^2).
If you did it again, three candies, you would have to have the above set for each of the three children again for the new candy.
AAA AAB AAC
BAA BAB BAC
CAA CAB CAC
ABA ABB ABC
BBA BBB BBC
CBA CBB CBC
ACA ACB ACC
BCA BCB BCC
CCA CCB CCC
Now there are 27 possibilities... that is 3(number of kids)^3(number of candies)
So there would be 3^9 (19,683) possibilities.

However, in light of Kerunpav's comment I would have to agree. There is no way... they will start eating them before you can hand them all out. =D

Thanks for playing... I'll post again next week if anyone is interested?

Nii · Oct 05, 2007, 12:21 AM // 00:21

Quote:

Originally Posted by Yellow_lid

Right idea, but you didn't quite do it correctly...
Each candy can only go to one child at a time... we shall call them children A, B, and C.
If you only had one candy there would be 3 possibilities (3^1).
A
B
C
If you had two candies how many would there be.
AA
BA
CA
AB
BB
CB
AC
BC
CC
now there are 9 possibilities... notice that I had to to square the number of possibilities when I threw a new candy into the mix (3^2).
If you did it again, three candies, you would have to have the above set for each of the three children again for the new candy.
AAA AAB AAC
BAA BAB BAC
CAA CAB CAC
ABA ABB ABC
BBA BBB BBC
CBA CBB CBC
ACA ACB ACC
BCA BCB BCC
CCA CCB CCC
Now there are 27 possibilities... that is 3(number of kids)^3(number of candies)
So there would be 3^9 (19,683) possibilities.

However, in light of Kerunpav's comment I would have to agree. There is no way... they will start eating them before you can hand them all out. =D

Thanks for playing... I'll post again next week if anyone is interested?

I don't get it Q(-.-Q)
AND yes im interested

Yellow_lid · Oct 05, 2007, 03:13 AM // 03:13

Quote:

Originally Posted by Nii

I don't get it Q(-.-Q)
AND yes im interested

It's quite simple... every time you add a candy in the above illustration there are 3x as many possibilities as the one before it.
If there was one candy
3=3^1=3
If there was 2
3x3=3^2=9
If there was 3
3x3x3=3^3=27
If there was 4
3x3x3x3=3^4=81
.
.
.
If there was 9
3x3x3x3x3x3x3x3x3=3^9=19,683

I'll have a new question posted by Monday.