Problematic Permutation Problem

Question

i see a problem without any definition. would you please help me?

i want to calculate the number of permutations of 1,2,...,1392 that 696 numbers be in the natural positions (from all numbers, 696 numbers be in the correct position).

i reach two solution but i don't know which one is correct?

a)

$$\frac{1392!}{696!}\sum_{k=0}^{696} \frac{(-1)^k}{k!}$$

b)

$$\binom{1392}{696}\sum_{k=0}^{696} \frac{(-1)^k}{k!}$$

Answer 1

It is much easier to help you if you show the reasoning behind your answers. b) is almost correct-first you choose the elements which will be fixed, then you choose a derangement of the rest, which makes sure none of them stay in place, but you missed a factor of $696!$. That will get you to a)

Answer 2

There are $\binom{1392}{696}$ ways to choose the fixed points, then $$696!\sum_{k=0}^{696} \frac{(-1)^k}{k!}$$ derangements, so (a) is the correct answer.

$$\binom{1392}{696}\cdot 696! = \frac{1392!}{696!}$$ so $(a)$ is correct.

Answer 3

You could check your answers and spot the error in one of them by looking at a simpler version with $4$ and $2$ rather than $1392$ and $696$.

So which of $\displaystyle \frac{4!}{2!}\sum_{k=0}^{2} \frac{(-1)^k}{k!}$ and $\displaystyle {4 \choose 2}\sum_{k=0}^{2} \frac{(-1)^k}{k!}$ gives a result of $6$?