selecting 4 non-consecutive books from 10 books. 
I have a set a $10$ book kept in a line and I want to find out how many ways $4$ books can be chosen from that if I don't choose consecutive books to be taken out.

I felt this is similar to forming 4 letter word not chosen consecutively from a 10 letter word. So the answer will be $10\times 8\times 6\times 4 = 1920$ ways. The possibility of first letter will be 10 and we omit the next letter so the possibility of second alphabet will be 8 and similarly it goes on. Is this correct?
 A: The same as the number of ways $4$ books can be inserted in gaps between the other six.
$\uparrow B \uparrow B \uparrow B \uparrow B \uparrow B \uparrow B \uparrow\;\; viz\;\; \dbinom74$
A: Write down $6$ stars, to represent in the abstract books not taken. 
$$\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast$$
These determine $7$ gaps, of which $5$ are interstar gaps, and the other two endgaps.
From these gaps, $4$ were selected for book removal. This can be done in $\binom{7}{4}$ ways.
A: Hint: 
Find the number of ways you can select $4$ consecutive books, $6$ ways is possible.
A: This is small enough that you can draw it out using placeholders such as this:
$1$ _ $3$ _ $5$ _ $7$..$10$ ($4$ choices)
$1$ _ $3$ _ _ $6$ _ $8$..$10$ ($3$ choices)
$1$ _ $3$ _ _ _ $7$ _ $9$..$10$ ($2$ choices)
$1$ _ $3$ _ _ _ _ $8$ _ $10$       (only $1$ choice)
$1$ _ _ $4$ _ $6$ _ $8$..$10$ ($3$ choices)
$1$ _ _ $4$ _ _ $7$ _ $9$..$10$ ($2$ choices)
$1$ _ _ $4$ _ _ _ $8$ _ $10$       (only $1$ choice)
$1$ _ _ _ $5$ _ $7$ _ $9$..$10$ ($2$ choices)
$1$ _ _ _ $5$ _ _ $8$ _ $10$ ($1$ choices)
$1$ _ _ _ _ $6$ _ $8$ _ $10$       (only $1$ choice)
_ $2$ _ $4$ _ $6$ _ $8$..$10$ ($3$ choices)
_ $2$ _ $4$ _ _ $7$ _ $9$..$10$ ($2$ choices)
_ $2$ _ $4$ _ _ _ $8$ _ $10$       (only $1$ choice)
_ $2$ _ _ $5$ _ $7$ _ $9$..$10$ ($2$ choices)
_ $2$ _ _ $5$ _ _ $8$ _ $10$ (only $1$ choice)
_ $2$ _ _ _ $6$ _ $8$ _ $10$       (only $1$ choice)  
_ _ $3$ _ $5$ _ $7$ _ $9$..$10$ ($2$ choices)
_ _ $3$ _ $5$ _ _ _ $8$ _ $10$       (only $1$ choice)
_ _ $3$ _ _ $6$ _ $8$ _ $10$ (only $1$ choice)  
The last will be:
_ _ _ $4$ _ $6$ _ $8$ _ $10$       (only $1$ choice)  
So $35$ ways possible.
