How many different sums can be formed with the following coins:
5 rupee, 1 rupee, 50 paisa, 25 paisa , 10 paisa,3 paisa, 2 paisa,1 paisa?

• Here we have 8 coins.
  We can select either 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 coins.
  so, different ways of selecting coins are=
  8c1+8c2+8c3+8c4+8c5+8c6+8c7+8c8=255
  we can use the shortcut formula for finding total no. of combination i.e 2^n-1
  =2^8-1=255.
  Then, we have one pair
  (1,2) which will give the sum 3 and it is already included in 3 paise coin.so, we have to remove it.
  so, the final ans will be,
  255-1=254.
• 7 years agoHelpfull: Yes(15) No(3)
• The answer should be 2^8 -2 i subtract 1 one more time because 3 paisa=2 paisa+1 paisa
• 7 years agoHelpfull: Yes(6) No(3)
• Total, we have 8 coins.
  Hence 2^8-1=256-1=255 possible combinations.
  Now, you might get same sum with different combinations.
  
  Like (1,2) and 3
  Also, (35,10,2,3) and 50
  And, (50,35,10,2,3) and 1
  
  For the 1st case, we have 2^5 combos. So, 64 combinations give 32 values.
  
  For the 2nd case, we have 2^3 combos. So, 16 combos give 8 values.
  
  For the last case, we have 4 combos.
  
  Finally, total 255-32-8-4=211 sums
• 7 years agoHelpfull: Yes(4) No(5)
• Here we have 8 coins.
  We can select either 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 coins.
  so, different ways of selecting coins are=
  8c1+8c2+8c3+8c4+8c5+8c6+8c7+8c8=255
  
  Then we have 3paisa=1 paisa+ 2paisa
  i.e we have double counted some values and we need to remove that
  explanation:In 8c2 we must have counted 5rs 3paisa,1rs 3paisa, 53 paisa, 28 paisa, 13 paisa and in 8c3 we counted same in following manner : 5rs (2paisa +1paisa ) similarly 1rs (2paisa +1paisa) etc.
  so clearly those sums are not different.
  Now we need we subtract 5c1+5c2+5c3+5c4+5c5 from 255 because double counted sum is caused when we took some coins from first 5 coins and then took either 3 paisa or (2 paisa + 1 paisa). so final answer would be 255-(5+10+10+5+1)=224
• 6 years agoHelpfull: Yes(4) No(0)
• I think it should be 8c1+8c2+8c3+8c4+8c5+8c6+8c7+8c8=2^8-1=255
• 7 years agoHelpfull: Yes(3) No(11)
• question from arun sharma example num:17.13
• 7 years agoHelpfull: Yes(3) No(2)
• See, we have 6 different domains to add up and form a particular sum.
  
  =>Now as according to PnC, total number of selection (with zero) out of n different things is= nC0+nC1+nC2+nC3+...+nCn= 2^n.
  
  =>So for total non zero selection it will be= 2^n-1.
  
  hence here n=8
  
  hence number of sum possible= (2^8)-1 =255.
• 7 years agoHelpfull: Yes(1) No(1)
• check out this link.It will help
  https://gateoverflow.in/217975/%23permuatation?show=218002
• 6 years agoHelpfull: Yes(0) No(1)