SOLUTION: 5 boys and 3 girls are to be seated on chairs arranged in a row. If the arrangement is made by a random selection, find the probability that no two girls are selected side by side.

We only need to consider arrangements by gender only.
  
We take the 3 girls and 2 boys to set up the minimum requirements
for a success: 

__GB__GB__G__

Now we have to place the remaining 3 boys in the 4 groups represented
by the 4 blanks.  This is the number of ordered partitions of 3 things 
into 4 groups allowing groups of size 0.

The number of ordered partitions of n things into r groups is C(n+r-1,r-1), 
counting groups of 0.

So we have n=3, r=4

C(3+4-1,4-1) = C(6,3) = 20 ways to place the remaining boys into those
4 blanks (including leaving some of the blanks empty).

Just for illustrative purposes, those 20 arrangements by gender only are:

1.  GBGBGBBB
2.  GBGBBGBB
3.  GBGBBBGB
4.  GBGBBBBG
5.  GBBGBGBB
6.  GBBGBBGB
7.  GBBGBBBG
8.  GBBBGBGB
9.  GBBBGBBG
10.  GBBBBGBG
11.  BGBGBGBB
12.  BGBGBBGB
13.  BGBGBBBG
14.  BGBBGBGB
15.  BGBBGBBG
16.  BGBBBGBG
17.  BBGBGBGB
18.  BBGBGBBG
19.  BBGBBGBG
20.  BBBGBGBG

So the numerator of the desired probability is 20.

The denominator is the total number of gender arrangements.

We can pick any 3 of the 8 positions to place the 3 girls 
C(8,3) = 56 ways 

Answer C(6,3)/C(8,3) = 20/56 = 5/14

Edwin