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


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 ? 
