Question

p

Answer 1

\[(2^3-1) \times (2^4-1) \times (2^5-1) \]

Answer 2

???

Answer 3

how did you put up that combination??

Answer 4

Dinner time here. Let the OP concur the answer first :)

Answer 5

hmmm.... I'm confused, I would have thought it would be:\[3*4*5\]
if we let a=apple, b=banana, m=mango, then we get:
a, b, m
a, b, 2m
a, b, 3m
a, b, 4m
a, b, 5m

a, 2b, m
a, 2b, 2m
a, 2b, 3m
a, 2b, 4m
a, 2b, 5m

a, 3b, m
a, 3b, 2m
a, 3b, 3m
a, 3b, 4m
a, 3b, 5m

a, 4b, m
a, 4b, 2m
a, 4b, 3m
a, 4b, 4m
a, 4b, 5m

2a, b, m
2a, b, 2m
2a, b, 3m
2a, b, 4m
2a, b, 5m

etc...

Answer 6

Mea culpa! :( I have assumed the fruits to be distinguishable and at-least one fruit of each type to be chosen.

Answer 7

More Latin to learn! :D

Answer 8

Heh :D

Answer 9

Reading the question carefully, I think the answer should be \[ (3+1) \times (4+1) \times (5+1) -1\]

@Zarkon

Answer 10

I thought
\[ \sum_{i,j,k=1}^{3,4,5}C(3,i)*C(4,j)*C(5,k)\]