Here is my solution.
Let's start by assigning variables.
r is the number of red marbles, w is the number of white marbles, b is the number of blue marbles and g is the number of green marbles.
Moving on, there are r*w*b*g ways to choose one marble of each color.
There are \(wb\binom{r}2\) ways to choose one white, one blue and two reds.
Then, there are \(b\binom{r}3\) to choose one blue and three reds.
Finally, there are \(\binom{r}4\) ways to choose four reds.
The problem tells us that these are all equal, so
\(rwbg=wb\binom{r}2=b\binom{r}3=\binom{r}4\)
The first equality gives us 2g=r-1 which is the same as r=2g+1.
The second equality gives us 3w=r-2 which is the same as r=3w+2.
The final equality gives us 4b=r-3 which is the same as r=4b+3.
Looking at this, we can conclude that r is one less than a multiple of two, three and four.
The smallest number that fits all these categories is 11, and since we want the smallest amount possible, that is $r$.
Now we can just substitute r in the equalities we got.
11=2g+1 gives us g=5.
11=3w+2 gives us w=3.
11=4b+1 gives us b=2.
Therefore, the smallest number of marbles meeting the requirements is 11+5+3+2=21 marbles
