If a traveller was offered 5 destinations in asia and 11 destinations in europe and asked to chose utmost 3 destinations in asia and overall a 11 destinations. How many ways can he travel

• 5c3*11c8+5c2*11c9+5c1*11c10+1=2256
  
• 10 years agoHelpfull: Yes(22) No(1)
• He can choose atmost 3 out of 5 destinations in Asia and 11 destinations overall.
  Also it is mentioned that there are 11 destinations in Europe:
  No. of ways this can be done is :-
  5c3*11c8+5c2*11c9+5c1*11c10+5c0*11c11= 2256 ways
• 9 years agoHelpfull: Yes(8) No(0)
• ans is 2256
  5C0*11C11 + 5c1*11c10 + 5c2*11c9 + 5c3*11c8 = 2256
• 10 years agoHelpfull: Yes(5) No(0)
• if it is atmost then it will be 5c3*11c8
  and if it is atleast then it will be
  5c3*11c8+5c2*11c9+5c1*11c10+1=2256
• 10 years agoHelpfull: Yes(3) No(0)
• at least means kum se kum...
  so, according to condition we have to select 11 city in which at least 3 city from asia
  combinations are : 5c3*11c8 + 5c4*11c7 + 5c5*11c6
  1650+1650+462 = 3762
• 9 years agoHelpfull: Yes(2) No(1)
• =5c3*11c8+5c2*11c9+5c1*11c10+5c0*11c11
  =1650+550+55+1
  =2256
  
• 9 years agoHelpfull: Yes(2) No(0)
• 5c3*11c8+5c2*11c9+5c1*11c10+5c0*11c11= 2256 ways
• 9 years agoHelpfull: Yes(1) No(0)
• 2256 is the ans
• 10 years agoHelpfull: Yes(0) No(0)
• 5C3*11C8+5C4*11C7+5C5*11C6
• 10 years agoHelpfull: Yes(0) No(0)
• atmost means max 3 destinations of asia a traveller can select and not more than tat,,so ans wil be...
  5c3*11c8+5c2*11c9+5c1*11c10+1=2256
• 10 years agoHelpfull: Yes(0) No(0)
• ans is 2256 as 5c3*11c8+5c2*11c9+5c1*11c10+1=2256
  The traveller can chose utmost 3 destination from asia so he can either chose
  1 station from asia and 10 stations from europe (5C1*11C10) OR
  2 stations from asia and 9 stations from europe (5C2*11C9) OR
  3 stations from asia and 8 station from europe (5C3*11C8)
• 9 years agoHelpfull: Yes(0) No(0)
• 3762 as by combination
• 9 years agoHelpfull: Yes(0) No(1)
• destination in asia=5
  destination in europe=11
  
  only 3 destination should be selected so the ways are ..
  5c3*11c8 + 5c4*11c7 + 5c5*11c6 is required Ans
• 9 years agoHelpfull: Yes(0) No(1)
• ans is 5^7=78125.
  because the last two digits should be 12,24,32,44,52. acc to divisibility rule .
  so by fixing these digits at last , no. of ways for each are=5^6.
  but total cases are 5,so total no of ways are 5^7.
• 9 years agoHelpfull: Yes(0) No(0)
• 5c0*11c11+5c1*11c10+5c2*11c9+5c3*11c8
• 9 years agoHelpfull: Yes(0) No(0)