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
