How many arrangements are there of the letters in word with restriction How many arrangements are there of the letters in REPETITIVENESS such that
the Ss are consecutive AND the first I comes (somewhere) before the first E ?

What i though is by viewing ss as one letter , it would solve the first restriction . 

And i have no idea how to count the second restriction . 

there are 13 spaces considering that ss is one space 

there are 2 i and 4 e 

still cant figure this out :(
First treat I and E as the same letter. Then, once you have decided where in the final arrangement they go, how many ways are there to distribute Is and Es to those positions, keeping in mind that the first of them must be an I?
Count letters: 14 total- 4 E's, 2 I's, 2 T's, 2 S's, 1 R, 1 P, 1 V, 1 N

You are on the right track with making the S's one letter, and making therefore 13 available spaces instead of 14.

1) Make SS one letter- 1 way

2) Pick 6 spots for I's and E's-    C(13,6)

3) Put I in the first spot of those you just picked- 1 way

4) Arrange remaining I and E's into the 5 other spots-  5!/4!

5) Arrange other letters among the 7 remaining spaces, remembering to include SS as ONE letter-   7!/2!

Now just multiply those numbers together:

1* C(13,6)* 1* 5!/4!* 7!/2!