You know what factorial is right? For example, 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1
You're given 8 \times 7 \times 6
That looks very similar to 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3\times 2\times 1
But notice that we're missing the ...5 \times 4 \times 3\times 2\times 1 i.e. the rest of the terms after 6.
And notice how 5 \times 4 \times 3\times 2\times 1 = 5!
So we can argue that \dfrac{8!}{5!} = \dfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3\times 2\times 1}{5 \times 4 \times 3\times 2\times 1}
That gets rid of all the terms after 6 so we're just left with \dfrac{8!}{5!} = 8 \times 7 \times 6
Now apply the same logic to 5 \times 4 \times 3
Spoiler
.
How could you get rid of all the terms after 3?
Once you have both, the answer should coalesce.
Thank you very much. I know it may seem simple but I struggle with this chapter in statistics so I appreciate your help. Have a good rest of your week!