there are 4 couple who have to cross a river,every husband has to go with his wife.find minimum ways in which they can cross the river,if there is only 1 boat

• 1>how many persons per ride at a time is not said ...may be dummy
  2> as no. of persons per ride is not specified then you may take a 8 persons on a single ride then answer is 1
  3>if you take 2 persons at a time(generally this is the condition)
  h1w1--->
  
  
  
  
  
  
• 10 years agoHelpfull: Yes(17) No(0)
• 16 ways
  H1 would go either with H2 or H3 or H4.
  similarly H2 can go with either H3 or H4
  and the H3 can go with H4.
  so total possibilities is 6ways.
  
  similarly for its 6ways
  and
  3rd case is H1 can go with W1, H2 --> W2, H3--> W3, H4 --> W4 which equals 4 ways.
  so total ways is 6+6+4= 16 ways.
• 10 years agoHelpfull: Yes(9) No(14)
• MM WW HW
  4C2 4c2 4ways
  6 + 6 + 4=16 ways
• 10 years agoHelpfull: Yes(6) No(3)
• its 7.......
• 10 years agoHelpfull: Yes(5) No(0)
• 15 ways..
  
• 10 years agoHelpfull: Yes(3) No(0)
• I think this may help:
  http://www.mathsisfun.com/puzzles/the-four-elopements-solution.html
• 10 years agoHelpfull: Yes(2) No(0)
• here the capacity of the boat is not given that how many people can travel on boat at a time..
  so no. of minimum way can be considered to be 1.
• 10 years agoHelpfull: Yes(1) No(0)
• if considering that we are taking a couple at a time . then simply it will be 4*3*2*1 = 24
• 10 years agoHelpfull: Yes(1) No(2)
• Dummy else i think total of 6 times,
  4 times for each couple,
  then 2 times for 2 man each !!!
• 10 years agoHelpfull: Yes(1) No(1)
• how many people can go at a time...?
• 10 years agoHelpfull: Yes(0) No(1)
• 1/24 is the answer
• 10 years agoHelpfull: Yes(0) No(4)
• ans 16
  (m1*m2)=m2, (m1*w1)=m2(m1*w1); then m2 take (m2*m3)=m3,(m2*w2)=m3(m2*w2); like wise
• 10 years agoHelpfull: Yes(0) No(0)