Question 1031493
Between any two consecutive -'s, there has to be at least one + in between.


THe number of such ways is *equal* to the number of ways we can arrange three +'s and four -'s with no restrictions: for any arrangement of 3 +'s and 4 -'s, I can append a + after the first three -'s like so:


+--+-+- becomes +-+-++-++-


And vice versa (formally, there is a bijection between length-10 strings with 6 +'s, 4 -'s, no two consecutive -'s, and length-7 strings with 3 +'s, 4 -'s).


The number of length-7 strings with 3 +'s, 4 -'s is 7C3 = 7!/(4!*3!) = 35.