Instinct to thrive

Beautiful problem, It can simply be done by symmetry telling that in first day , the expected number of cells on the second day is 3,

After that, each cell feels the same driving force and hence tends to divide in 3 parts, successively and hence answer is $3^{20}$ , but Raghav has already shown this way

But for the sake of fun, i want to show a more sophisticated method

Now each cell has 4 choices ,

1) Not divide with probability 1/10

2) divide into two parts with probability 2/10

3) Divide into 3 parts with probability 3/10

4) Divide into 4 parts with probability 4/10

Since the possibilities are not equally likely, so let us rather think of this in this manner,

There is 1 way to not divide at all, 2 ways to divide into two cells, and 3 ways to divide into 3 cells and 4 ways to divide into 4 cells

so there are 10 paths for each cell, some paths having identical results,

Let them be represented by the coefficient of $x^r$ and thus the choice of each cell can be represented as

$\left( x+{ 2x }^{ 2 }+3{ x }^{ 3 }+{ 4x }^{ 4 } \right)$

and if in some day, there are n cells,

Then the number of possibilities of getting p cells is the coefficient of $x^p$ in

$\left( x+{ 2x }^{ 2 }+3{ x }^{ 3 }+{ 4x }^{ 4 } \right) ^{ n }$

Now what we want is the expectation of next day,

and expectation is defined as as the sum of probabilities with their respective values

or

since minimum number of cells next day is n and max is 4n

So the expectation is

${ P }_{ n }n+{ P }_{ n+1 }(n+1)\quad +\quad ....\quad { P }_{ 4n }(4n)$

To find that let $\left( x+{ 2x }^{ 2 }+3{ x }^{ 3 }+{ 4x }^{ 4 } \right) ^{ n }$ = $a{ x }+b{ x }^{ 2 }+c{ x }^{ 3 }.....\quad$

since the coefficients simply represent the number of ways of the particular exponent (cell no.)

So let us differentiate it to get

$a+2bx+3c{ x }^{ 2 }....\quad =\quad (n\left( x+{ 2x }^{ 2 }+3{ x }^{ 3 }+{ 4x }^{ 4 } \right) ^{ n-1 })(1+4x+9{ x }^{ 2 }+16{ x }^{ 3 })$

Now putting x=1 we get

E*(total number of possiblities) = $n{ 10 }^{ n-1 }(30)$

and each cell has 10 choices, so total number of possibilties is simply $10^n$

so E = 3n

where n is the number of cells in the previous day, and E is the expected number of cells the next day,

This recurrence is simply a GP

and initially there was just one cell

so after 20 days

${ E }_{ 20 }\quad =\quad { 3 }^{ 20 }$

Which is the answer,

I know the method is really big and unnecessary, but at the same time its fun

Let's see here:

At the beginning of day 1, there is one cell. Let's say the expected(average) value of population on the $n^{th}$ day is $E[n]$ .

At the start of the second day, we have a few daughter cells, which behave EXACTLY the same way as the parent. That is, $E[]$ functions for each daughter is same as the $E[]$ function of the parent, but shifted by one day.

Hence:

$E[n]=\sum _{ r=1 }^{ 4 }{ \frac { { r }^{ 2 } }{ (\frac { 4 }{ 2 } )(5) } } E[n-1]\\ \Rightarrow E[n]=(\frac { 9 }{ 3 } )E[n-1]\\ \Rightarrow E[n]={ (3) }^{ n-1 }E[1]={ (3) }^{ n-1 }$

Putting $n=21$ yields answer as $3^{20}=3486784401$