1. ## Combinations

I am having trouble with this question.

2. Originally Posted by Stuck Man
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..

$\binom{10}{10}=1$

In how many ways can you correctly forecast only one result?
You must correctly forecast one of the 10 results..

$\binom{10}{1}=10$

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).

3. Its 120. I have no idea what to do next.

4. Originally Posted by Stuck Man
Its 120. I have no idea what to do next.
As written, that's all that's required.
If that's the answer in the book, then you need to know how to get it,
but if you worked it out, then that's the question's answer.

5. Following from what you said I found the value of 10C7. The answer in the book is 3240.

6. I guess they are taking into account the fact that there can be 4 possible results for each game after all.

7. Using the binomial expansion

$(x+y)^n=\binom{n}{0}x^0y^n+\binom{n}{1}x^{1}y^{n-1}+....$

$p=0.25,\;\;\; q=0.75$

where p is the probability of a correct forecast,
the probability of getting exactly 7 results correctly forecasted is $\binom{10}{7}0.25^7\, 0.75^3$

The number of possible forecasts is $4^{10}$

Multiplying these gives you the result
(I thought there might be a catch to it).