I am having trouble with this question.
In how many ways can you get all 10 results correctly forecasted?
You must choose all results correctly in advance.
There's only one way you can do that..
$\displaystyle \binom{10}{10}=1$
In how many ways can you correctly forecast only one result?
You must correctly forecast one of the 10 results..
$\displaystyle \binom{10}{1}=10$
In how many ways can you correctly forecast exactly 7 correct results?
The fact that any game can have one of four possible "results" does not come into this (yet).
Using the binomial expansion
$\displaystyle (x+y)^n=\binom{n}{0}x^0y^n+\binom{n}{1}x^{1}y^{n-1}+....$
$\displaystyle p=0.25,\;\;\; q=0.75$
where p is the probability of a correct forecast,
the probability of getting exactly 7 results correctly forecasted is $\displaystyle \binom{10}{7}0.25^7\, 0.75^3$
The number of possible forecasts is $\displaystyle 4^{10}$
Multiplying these gives you the result
(I thought there might be a catch to it).