I am looking at the following sum $$ \sum c_1\wedge \cdots\wedge c_n $$ where the summation ranges over $c_1,\ldots,c_n$ such that each $c_i\in\{a,b\}$ and $a$ appears exactly $j$ times. Thus, using the definition of the exterior product, for such $c_i$'s one can always write $$ c_1\wedge \cdots \wedge c_n= \pm \;\underbrace{a\wedge\cdots\wedge a}_j\wedge \underbrace{b\wedge \cdots\wedge b}_{n-j}. $$
Is there an explicit formula for the choice of $+$ or $-$ ?
Is there a closed formula for ${\mathscr Coef}(n,j)$ defined by $$ \sum c_1\wedge \cdots\wedge c_n = {\mathscr Coef} (n,j) \; \underbrace{a\wedge\cdots\wedge a}_j\wedge \underbrace{b\wedge \cdots\wedge b}_{n-j}\qquad ? $$
I understand this problem as "you have an ordered configuration of $n$ marbles, each having color black or white, and exactly $j$ of these are black. What is the signature of the permutation required to order them as all the black marbles, and then all the white ones ? What is the sum of the signatures over all possible configurations ?" But I can't get really far. Thanks in advance.
What you are asking about seems to be the sum over all $j$ elements subsets$~S$ of $\{1,2,\ldots,n\}$ of $(-1)^{i(S)}$ where $i(S)$ counts the number of inversions in$~S$: pairs of an element present in$~S$ and a larger element absent from$~S$. Equivalently the permutations of $j$ copies of the symbol $1$ and $n-j$ copies of the symbol$~0$ counted by the parity of the occurrences of $1\,0$, in that order but no necessarily adjacent.
If $n$ is even, divide the string into $n/2$ substrings of length$~2$. If any of them has unequal digits, swap them to find a cancelling string with opposite parity. What remains is the strings built out of copies of $00$ and $11$. Since these always have an even number of digits $1$, the answer is $0$ when $n$ is even but $j$ is odd. When both $n$ and $j$ are even, all remaining strings contribute $+1$, so the answer is $\binom{n/2}{j/2}$ for that case.
For $n$ odd, a simimlar argument shows the answer is $\binom{(n-1)/2}{\lfloor j/2\rfloor}$ (the last symbol must be chosen with the same parity as$~j$ to get any remaining contribution, and it is the same as what one gets without that symbol).