7! = 35r!(7-r)!

5040 = 35r!(7-r)!

Divide both sides by 35

144 = r!(7-r)!

2×2×2×2×3×3 = r!(7-r)!

The right is a product of 2 factorials. To have a solution 
we must be able to group the factors on the left to form the 
product of 2 factorials such that what they are the factorials 
of total to 7.

Since there are no factors of 5 or larger on the left
we can only be looking at factorials for r! and (7-r)!
of 4 or less.

We know that 3×2 = 3! so we can write the above as

(3×2)×2×2×2×3 = r!(7-r)!

   3!×2×2×2×3 = r!(7-r)!

Then we can write 2×2 as 4 and have

   3!×4×2×3 = r!(7-r)!

And since 4×2×3 = 4×3×2 = 4!,

       3!4! = r!(7-r)!

So we can have two possibilities:

3! = r! and 4! = (7-r)!
 3 = r  and  4 = 7-r
             r = 3

or we can have

4! = r! and 3! = (7-r)!
 4 = r  and  3 = 7-r
             r = 4

Two solutions: r = 3 and r = 4.

Edwin