How many different integers can be expressed as the sum of three distinct numbers from the set{13,10,23,28,33,36,43,48}?

• 8C3=56
  
  and
  10+23+48 gives an integer which is also obtain from 10+28+43
  .Similarly 10+23+33 can also b obtain from 10+13+43.
  Hence 56-2=54
• 9 years agoHelpfull: Yes(24) No(2)
• 8C3=8!/5!*3!=56
• 9 years agoHelpfull: Yes(20) No(12)
• please some0ne provide the correct solution
• 9 years agoHelpfull: Yes(2) No(0)
• ans is 54
  8C3=56
  56-2=54
• 9 years agoHelpfull: Yes(2) No(1)
• plz explain this
• 9 years agoHelpfull: Yes(1) No(0)
• total number of selection of 3 number out of 8 number is=8c3=56
  10+28+33=10+13+48
  10+13+43=10+23+33
  the answer is=56-2=54
• 9 years agoHelpfull: Yes(1) No(0)
• total possible integers=8C3. but there are 6 integers repeating twice. so we have to exclude 6. so total different integers are 56-6=50.
• 9 years agoHelpfull: Yes(1) No(1)
• Answer is 32.
  
  
• 9 years agoHelpfull: Yes(0) No(5)
• please give a proper valid reason for the answer 54?
  
• 9 years agoHelpfull: Yes(0) No(0)
• plzz explain it properly.. why 56-2?
• 9 years agoHelpfull: Yes(0) No(0)
• its not termed as all the no i belong to the set so the no is (0to9)
  so there is 2 option the sum is either 2 digit no. or 3 digit no.
  for the 2 digit no 9c1*9c1=81
  for the 3 digit no 9c1*9c2=324
  so there are 324+81=405 distinct no possible
  
• 9 years agoHelpfull: Yes(0) No(2)
• answer will be 54 because by 8C3 be choose set of intergers but out of this (13,10,23) & (23,10,33) are similar so for different set we get 56-2 that is 54
• 9 years agoHelpfull: Yes(0) No(0)
• 15 is the sol since the comm diff is 7 their sum also has a comm diff 7the min sium is 30 and max sum is 135 the remaining sums are multiple of 7 in between 30 and 135 . there fore the ans is (135-30)/7 = 105/7 = 15
• 8 years agoHelpfull: Yes(0) No(0)