Distribute balls to cells problem I couldn`t understand how did they get to (b) solution.
Can someone please give an explanation?
Thanks!

 A: If you have exatly two empty cells and one cell with three balls, the remaining four cells must contain one ball each.
$\dfrac{7!}{2!1!4!}$ counts ways to select two cells to be empty, one cell to hold three balls, and four cells to hold one ball each.
$\dbinom 7 3$ is the ways to select three balls from the seven to put into the three-balls cell.
$4!$ is the ways to arrange the remaining balls in one-ball cells.
And $7^7$ is the size of our sample space (count of all ways to select a cell for each ball).
Hence the probability measure is : $$\dfrac{7!}{2!1!4!}\dbinom 7 3 \frac{4!}{7^7}$$
Now, can you see how the other term is obtained?  

 Count ways to select two cells to be empty, two cells to hold two balls each, and three cells for one ball each, then select balls for those cells.

A: An explanation of the expression has already been given, I'd only like to add that since you are likely to encounter many problems of the balls in cells type,(and worked out differently in different books !), you might like to standardize the method.
The one I use for counting the arrangements is a multiplication of two multinomial coefficents, one for the number of balls in each cell, and the other for the frequency with which nulls, singles, doubles,... occur, e.g. for the two patterns possible here, viz.
$3-1-1-1-1-0-0: \binom7{3,1,1,1,1}\binom7{1,4,2}$ or the equivalent expression $\frac{7!}{3!1!1!1!1!}\cdot\frac{7!}{1!4!2!}$
$2-2-1-1-1-0-0:\binom7{2,2,1,1,1}\binom7{2,3,2}$ or the equivalent expression $\frac{7!}{2!2!1!1!1!}\cdot\frac{7!}{2!3!2!}$
Add up, and divide by the sample space $7^7$
This way the process requires least thought, and is less error prone.

NOTE
Talking of error proneness, the part of the book's expression that reads
$\frac{7!}{2!2!3!1!1!}$ is wrong !Can you tell why ? 
