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The number of binary strings of $n$ zeros and $k$ ones in which no two ones are adjacent is

  1. $^{n-1}C_k$
  2. $^nC_k$
  3. $^nC_{k+1}$
  4. None of the above
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5 Answers

Best answer
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63 votes

Answer is (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  $=^{n+1}C_k$

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$\text{i think n$\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}$

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Total no of ways = C(n+1 ,k)

The correct answer is (D) None of the above

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0 votes

Option D, None of these

As every 0 is identical we can place n 0's in 1 way only. Placing of n 0's will create n+1 slots where we can place 1's as to satisfy the given condition & moreover as every 1 is also identical they'll also be arranged in 1 way only.

So, Total number of binary strings possible is

1*(n+1)Ck *1  = (n+1)Ck 

Answer:

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