FIFTY POINTS!! Please help! I'll give brainliest A school of 100 fish swims in the ocean and comes to a very wide horizontal pipe. The fish have three choices to get to the food on the other side: swim above the pipe, through the pipe or below the pipe. If we do not consider the fish individually, in how many ways can the entire school of fish be partitioned into three groups with each group choosing a different one of the three options and with at least one fish in each group?

4851 ways

Step-by-step explanation:

The fish have 3 choices. They can make it above, below, or though the pipe. Keep in mind there are 100 fish total:

group 1 + group 2 + group 3 = 100

If we keep group 3 (the fish that swim below the pipe) constant, say 1, and increment the other two (group 2 starting off at 1) we find 98 possibilities.

   98     +      1        +      1        = 100,

   97      +     2        +      1        = 100,

   96      +     3        +      1       =  100

        .           .            . 98 possibilities

Now we take group 1 as one greater (1 + 1 = 2) and then start incrementing group 2 starting from 1 as done before. So 97 + 2 + 1 = 100. Followed by 96 + 3 + 1, 95 + 4 + 1...97 possibilities

If we continue this pattern, we have 98 + 97 + 96... + 1 total possibilities to partition this school of fish.

98 + 97 + 96... + 1,

Sum = n(n + 1)/2 = 98(98 + 1)/2 = 98(99)/2 = 4851

The simplified expression would be 4

Step-by-step explanation:

So since we have parentheses in this expression we have to use the steps of PEMDAS (parentheses, Exponent, Multiplication, Division, Addition, Subtraction)

So the first step is to solve what's in the parentheses

Then we multiply

Since we have simplified the equation so far let's write it out

Now we just solve the equation as written