## $\text{i think n$\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}$