The number of binary strings of $n$ zeros and $k$ ones in which no two ones are adjacent is
$^{n-1}C_k$
$^nC_k$
$^nC_{k+1}$
None of the above
answer - D
first place n zeroes side by side _ 0 _ 0 _ 0 ... 0 _
k 1's can be placed in any of the (n+1) available gaps hence number of ways = ^{n+1}C_{k}
Shouldn't the complete answer be n+1Ck multiplied by k! multiplied by (k-1)!.taking into consideration the arrangements of 0 s
Arrangement of n zeroes can be in Ways as
_0_0_....._0_
Keeping n+1 places for ones so that no two ones can be placed together and ther are k ones to be placed.
= ^{n+1}C_{k}
all 0s and 1s are identical so no need to permutate them
Total no of ways = C(n+1 ,k)
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