$\text{i think n$\geq$k-1}$
$\text{then we will have $^{n+1}C_k$}$
$\text{suppose we have 3 zeroes and 2 1's now n+1 gives 4 places after placing}$
$\text{zeroes in this way(_0_0_0_) 0's can be placed in one way now}$
$\text{,from the 4 places we can select 2}$
$\text{places and can put the two 1's it can look this ->10100 or 01010 $\cdot$$\cdot$}$
$\text{note->no need to arrange the zeroes and 1's as they are identical if}$
$\text{you will arrange then it will be like this}$
$\text{$^{3+1}C_2$$\frac{3!}{3!}\frac{2!}{2!}$}$
$\mathbf{alternate approach }$
$\text{by option elimination,take n=2 (0's),k=2(1's) then it gives 1001,0101,1010}$
$\text{a) $^{1}C_2$ (wrong) b)$^{2}C_2=1$ (wrong) ,c)$^{2}C_3$ (wrong) d)correct}$