A row of seats in a movie hall contains 10 seats. 3 Girls

Are you sure?
My answer is 1/20.

I tell you why:
First, we have to identify the total number of combinations of seats in which the three girls can occupy.
So, 10C3 = 120 possible combinations.

Now, we have to identify the combinations in which there are not two girls together.

Only to offer a better explanation, I will write the combinations:

Seat N°: 1 2 3 4 5 6 7 8 9 10
G B G B G B B B B B
B G B G B G B B B B
B B G B G B G B B B
B B B G B G B G B B
B B B B G B G B G B
B B B B B G B G B G

Based on this, there are only 6 scenarios in which there are not two girls together.
Obviously, it is not necessary to write this combinations, you only have to see that when the third girl is in the 10th position, the first girl is in the 6th. But the first girl cannot be in the 7th, because there is not the 11th position for the third girl.
(I think that there is a way to solve this part with combinatronics, but I didn't find it)

So, we have this 6 scenarios and the total number of events (120).
We divide: 6 / 120 = 1/20

What do you think?

I think I deserve kudos

This approach is not correct. There are more cases possible, for no girls to sit together, for example:

G B B G B G B B B B
Or:
G B G B B B B B B G
...

Answer given by 4gmatmumbai is correct.

A row of seats in a movie hall contains 10 seats. 3 Girls & 7 boys need to occupy those seats. What is the probability that no two girls will sit together?

Consider the following:
*B*B*B*B*B*B*B*

Now, if girls will occupy the places of 8 stars no girl will sit together.

# of ways 3 girls can occupy the places of these 8 stars is \(C^3_8\);
# of ways 3 girls can be arranged on these places is \(3!\);
# of ways 7 boys can be arranged is \(7!\).

So total # of ways to arrange 3 Girls and 7 boys so that no girls are together is \(C^3_8*3!*7!\);
Total # of ways to arrange 10 children is \(10!\).

So \(P=\frac{C^3_8*3!*7!}{10!}=\frac{7}{15}\).

Hope it's clear.

Could you plz help me why my approach is wrong?

P(not 2 girls sit together)=1-P(2 girls sit together or 3 girls sit together)
If we name girls as X, Y, and Z then the above equation:

p(not 2 girls sit together)=1-p(XY or XZ or ZY or XYZ)

\(P(xy)=P(xz)=P(zy)= \frac{9!*2}{10!}\)

\(P(xyz)=\frac{8!*3!}{10!}\)

So, P(not 2 girls sit together)=\(1-\frac{9!*2+9!*2+9!*2+8!*3!}{10!}=1-\frac{2}{3}=\frac{1}{3}\)

What's wrong???

You didn't distinguish between the cases of exactly two girls sitting together and one sitting alone and the case when all three girls sit together.
\(P(xy)=P(xz)=P(zy)= \frac{9!*2}{10!}\) each of these also includes some cases when all three girls sit together.
Now, it gets even more complicated. When considering xy sitting together, it might happen that z joined them, so you also had the arrangement xyz or zxy, but since you took only 2! (just switching between x and y only), you subtracted only 4 out of the full 3!=6 possibilities. Which means that you subtracted \(3\cdot\frac{4}{6}=2\) or twice more the probability P(xyz).

To your result, you should add back twice P(xyz). You will get \(\frac{1}{3}+2\cdot\frac{2\cdot3}{9\cdot10}=\frac{1}{3}+\frac{2}{15}=\frac{7}{15}.\)