S sid_178 Sep 2008 14 0 May 14, 2010 #1 The number of binary strings of n zeroes and k ones that no two ones are adjacent are : (a)n-1Ck (b)n+1Ck (c)nCk d)k+n-1Ck

The number of binary strings of n zeroes and k ones that no two ones are adjacent are : (a)n-1Ck (b)n+1Ck (c)nCk d)k+n-1Ck

P Plato MHF Helper Aug 2006 22,462 8,634 May 14, 2010 #2 sid_178 said: The number of binary strings of n zeroes and k ones that no two ones are adjacent are : (a)n-1Ck (b)n+1Ck (c)nCk d)k+n-1Ck Click to expand... This is a good problem. What do you thin the answer is and why?

sid_178 said: The number of binary strings of n zeroes and k ones that no two ones are adjacent are : (a)n-1Ck (b)n+1Ck (c)nCk d)k+n-1Ck Click to expand... This is a good problem. What do you thin the answer is and why?

S sid_178 Sep 2008 14 0 May 14, 2010 #3 By Assumption ,say ,n>k.. Now ,if we keep n zeroes ..we have n+1 places out of which we can keep k ones in n+1ck ways .Is that right ?

By Assumption ,say ,n>k.. Now ,if we keep n zeroes ..we have n+1 places out of which we can keep k ones in n+1ck ways .Is that right ?

P Plato MHF Helper Aug 2006 22,462 8,634 May 14, 2010 #4 sid_178 said: By Assumption ,say ,n>k.. Now ,if we keep n zeroes ..we have n+1 places out of which we can keep k ones in n+1ck ways .Is that right ? Click to expand... Actually we assume that \(\displaystyle n\ge k-1\). Consider \(\displaystyle 10101\). So yes it is \(\displaystyle \binom{n+1}{k}\). Reactions: sid_178

sid_178 said: By Assumption ,say ,n>k.. Now ,if we keep n zeroes ..we have n+1 places out of which we can keep k ones in n+1ck ways .Is that right ? Click to expand... Actually we assume that \(\displaystyle n\ge k-1\). Consider \(\displaystyle 10101\). So yes it is \(\displaystyle \binom{n+1}{k}\).