How many ways are $3$ letters placed consecutively. ($6$ letters from $8$ letters) Here are the questions:


  $i)$How many ways can you arrange $6$ letters from the letters $a,b,c,d,e,f,g,h$? 
  
  $ii)$How many of those have $b,c$ and $g$ placed consecutively?


I think the solution to the first one is $8\choose 6$, I'm still not sure so please correct me if I'm wrong. I don't have an idea though how to answer the next question. Please help.
Thanks :)
The number of ways to choose 6 letters from the 8 is $\binom 8 6 = \frac{8!}{6! 2!}$ 

each of these choices can be arranged in $6!$ ways - so the answer to i) is $\frac{8!}{ 2!}$

for ii) - there are $3!$ ways to arrange b,c,g consecutively and  $\binom 5 3 = \frac{5!}{3! 2!}$ choices of the other three letters. For each of these choices there will be $4!$ ways to arrange the 3 letters and the b,c,g block.

$$N=   \frac{3! 4! 5!}{3! 2!} =  \frac{4! 5!}{ 2!}$$
For B , C , G to occur consecutively take them as one letter only . Like X={b,c,g} you are left with 5 letters and select any 5 of them by 5C5 and the 6th would be X . You can arrange the letters inside X by 3! Ways and the six letters(which include X) by 6! Ways hence the answer should be 6! X 3!