# Thread: 3 lines 9 wrenches

1. ## 3 lines 9 wrenches

3 lines are to divide up 9 wrenches evenly.

$\displaystyle\frac{9!}{(3!)^3}=1680$

I have this part correct but I can't seem to get this next part.

If there are 2 used and 7 new wrenches, what is the probability that a particular line (line A) gets both used?

2. Originally Posted by dwsmith
3 lines are to divide up 9 wrenches evenly.
What is the world does that statement mean?

3. Originally Posted by Plato
What is the world does that statement mean?
Assembly lines.

4. The answer $\dfrac{9!}{(3!)^3}$ is the number of ways to divide nine people, two adults and seven children, into three different activates of three each.
If we want the two adults in a particular same activity, that can be done is one way and seven ways to pick the one child.
How many ways to assign six children to two actives of three each?