NewHentai

we can use basic definition of probability to solve this

probability = successful outcomes / total outcomes

calculating total outcomes:

A has 3 options for group

B has 3 options for group

C has 3 options for group

so total possibilities = 3 * 3 * 3 = 27

now there are 3 total successful outcomes: One where all 3 go in first group one where all 3 go in 2nd group and one where all 3 go in 3rd group

so probability = 3/27 = \(\boxed{\frac{1}{9}} \)

ok we have the integers 0,1,2...,9

to make the sum equal to 4 we have options

4 + 0 = 4

3 + 1 = 4

2 + 2 = 4 (not possible as we cant have duplicates)

and that is all

so new we can divide this into cases.

case # 1 (4 and 0 are the largest 2 elements):

if 0 is second largest element we have no element less than it so the only subset possible is {0, 4} so 1 option

case # 2 (3 and 1 as 2 max elements):

now we obviously have 0 which we can add such that it does not mess with the max elements as any element >= 2 it would mess up the "second-largest = 1" so we have two options one with zero and one without zero {0,1,3}, {1,3}

so adding up the cases we have 3 satisfying subsets