There are N number of detectives , each one knows an information, how many calls should they make so all the detectives know all the N number of info. ?

• sorry the solution is ---> N(N-1)/2
• 11 years agoHelpfull: Yes(19) No(4)
• (n-1)+(n-2). !st one will call the 2nd and give information and take information. Then he will call the 3rd and will give the information of his and 2nd detective to him. this process continues till n-1 detective. Now he has information of all the detective. So he will call all other n-2 detective(except him and last one who already has all the information) and will give the information to all.
  so total call=(n-1)+(n-2)
• 11 years agoHelpfull: Yes(16) No(3)
• each one makes n-1 calls , so ans is n*n-1
• 11 years agoHelpfull: Yes(9) No(6)
• (n-1) ways
• 11 years agoHelpfull: Yes(3) No(6)
• The answer is (n-1)+(n-2)+...+1=(n(n-1))/2.
  First caller needs to make n-1 calls,
  Second caller needs to make n-2 calls since he knows the secret of the first caller and so on...
• 11 years agoHelpfull: Yes(3) No(0)
• Total Calls = C(N,2)= [N(N-1)]/2
  
  Explanation:
  1 call needed to share information between each of two detectives.
  If there are N detectives., it need C(N 2) combinations.
• 11 years agoHelpfull: Yes(2) No(1)
• sorry the answer is n-1! i.e n-1 factorial....
• 11 years agoHelpfull: Yes(2) No(3)
• 1st detective calls n-1 other detectives and gives them the information of all the detectives whom he has called just before calling a particular one.
  
  So, at the end of 1st cycle, 1st detective and last detective both will have N number of info. Detective n-1 will have n-1 info. Detective n-2 will have n-2 info and so on. 2nd detective will have 2 info.
  
  Now, another cycle starts in which nth detective calls detectives 2-(n-1) giving them info. of all the previously called detectives (same as in 1st cycle).
  
  Thus, at the end of this cycle, all the detectives will have N no. of info.
  
  So, total no. of calls =(N-1) + (N-2) = 2*N-3
• 10 years agoHelpfull: Yes(2) No(1)
• (n-1)+(n-2)=2n-3
  
• 10 years agoHelpfull: Yes(2) No(0)
• answer is
  nC2...
  
• 11 years agoHelpfull: Yes(1) No(2)
• nth det. has to make n-1 calls
  (n-1)th .... ..... ..... n-2 calls
  .
  .
  1st...........1 call
  total no of call=(n-1)+(n-2)+...+1
  =>(n-1)(n-2)/2
  
• 11 years agoHelpfull: Yes(1) No(0)
• selecting 2 out of n = nC2=n(n-1)/2
• 11 years agoHelpfull: Yes(1) No(0)
• (n-1) ways
  
• 11 years agoHelpfull: Yes(1) No(0)
• the answer is n(n-1)/2
• 10 years agoHelpfull: Yes(1) No(0)
• N-1:for gathering whole info
  
  N-2:Share the all information
  
  Total: (N-1)+(N-3)=2N-3
  
• 10 years agoHelpfull: Yes(1) No(0)
• numCall=0
  while(N!=1)
  {
  x += N/2;
  N -= N/2;
  }
• 11 years agoHelpfull: Yes(0) No(4)
• (N(N+1)/2)-1
• 11 years agoHelpfull: Yes(0) No(0)
• see,each detective will call n-1 no of detectives to inform them about his information...so n detectives will make total n*(n-1) no of calls.
• 11 years agoHelpfull: Yes(0) No(0)
• n(n-1) or nC2 * 2!
• 11 years agoHelpfull: Yes(0) No(0)
• n-1 first one will call second..second will call third telling info of 1st one too..so each call next one..n-1 ways
• 11 years agoHelpfull: Yes(0) No(0)
• As there are n persons . each of them has to make n-1 calls to inform of his/her piece of information to others as he/she will exclude himself/herself for his/her information as he/she already knows it . so if there are 2 persons the number of calls to be made is 2*(n-1). And hence, for n number of persons it should be n*(n-1).
• 11 years agoHelpfull: Yes(0) No(0)
• Each detective knows only one information.
  so, there are n no. of information. so, they have to know n-1 more information. so, n-1 call is needed to know ALL the information of a single detective.. thre are n-1 more detectives so,, total no. of calls-->> n*(n-1)
• 11 years agoHelpfull: Yes(0) No(0)
• it is (n-1)! ways.
  
• 10 years agoHelpfull: Yes(0) No(0)
• If in the given question we assume that the info sharing is happening two-way when a call is made then ans= [n*(n-1)]/2 ......but if the sharing of info is just one-way when a call is made then the ans= [n*(n-1)]
• 10 years agoHelpfull: Yes(0) No(0)
• N=information
  N-1 call to share the all information.
• 10 years agoHelpfull: Yes(0) No(0)
• N-1
  if suppose there are three detectives.one call the second and give him the information.now second calls third and give him the information of himself and the first detectives.on similar grounds it would take N-1 calls
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• N^2
  each one will make n calls as the known info may be repeated
• 10 years agoHelpfull: Yes(0) No(0)
• 1st detective calls N-1 others
  then 2nd detective calls 1st one and can collect all the info.=1 call
  so on for N-1 detectives
  total calls= N-1+N-1 = 2N-2
• 10 years agoHelpfull: Yes(0) No(0)
• 2(n-1)
  1st detective will call 2nd one and share al his information then, 2nd detective will call 3rd one and he will share his information as well as information from 1st detective. And this process will continue till the nth detective obtain all the information. so till now there will be (n-1) calls
  after that nth detective will call 1st detective to share all the information and then 1st one will call 2nd one and 2nd one will call the 3rd one and this process will continue till the (n-1)th detective obtain all the information. so here also (n-1) call were made.
  so total calls made= (n-1) + (n-1)
  = 2n-2= 2(n-1)
• 10 years agoHelpfull: Yes(0) No(0)
• n*n-n
  because each one will make to others all and we need to substract n because they will exchange of their own. so substract the n
• 10 years agoHelpfull: Yes(0) No(0)
• (n-1)!
  lets say 4 detectives.
  N1 -> N2
  N1 -> N3
  N1 -> N4
  
  3 CALLS (N1) + 2 CALLS (NOW BY N2)+ 1 CALL (NOW BY N3) i.e. (n-1)!
• 10 years agoHelpfull: Yes(0) No(0)
• n/2(log n)
  (1,2,3,4,5,6,7,8) =4
  /
  (1,2,3,4) (5,6,7,8) =4
  / /
  (1,2) (3,4) (5,6) (7,8) =4
  n/2 at each level. If any information is exchanged by any ditective then both will contain the union of information.
• 7 years agoHelpfull: Yes(0) No(1)
• one should make n-1 calls excluding him.For n members,the calls should be n*n-1
• 7 years agoHelpfull: Yes(0) No(0)