Combinatorial Question (Numbers of combinations)

[ych22]

Homework Statement

In how many ways can 12 balls be arranged into 4 different rows with each row having at least one ball if there are 6 identical red balls and 6 identical white balls?

Homework Equations

number of combinations of N objects into r groups is $$\frac{(N+r-1)!}{N!(r-1)!}$$.

The Attempt at a Solution

I thought of two directions
1) First number of arrangements the 12 balls into 4 rows without the requirement of minimum of one ball in each row. Then find the number of combinations in which the requirement is not met. Deduct the latter from the former.

2) Pick the first ball in each row first. Then find the number of combinations to distribute the remaining balls.

Either way, I'm stuck on the math! I think I'm missing a deceptively easy way to look at the problem. Any hints would be appreciated.

 

[lanedance]

i would first assume you have 12 balls, and deal with the colours later

consider the 12 balls as follows
xxxxxxxxxxxx

partitioning them is the same as choosing 4 partitions from 11 available spots, example is given below
x|xxx|xx|xx|xxxxx

this will also guarantee there is at least one ball in each row, now include the different possible colour combinations