Latest Elitmus Aptitude Question SOLUTION: 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?

• Placement Papers
• /
• Elitmus
• /
• Numerical Ability
• /
• Permutation and Combination

Elitmus Question Papers:

(#M40164996) ELITMUS Numerical Ability question Permutation and combination Keep an EYE

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?

Asked In Elitmus Abhay katiyar (6 years ago)

Unsolved Read Solution (8)

Is this Puzzle helpful?   (18)   (9) Submit Your Solution Permutation and Combination

Read Solution (8) :

Akshma Mittal  (6 years ago)

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.

Is this solution Helpfull? Yes (15) | No (3)  

purushotam  (6 years ago)

The answer should be 2^8 -2 i subtract 1 one more time because 3 paisa=2 paisa+1 paisa

Is this solution Helpfull? Yes (6) | No (3)  

NITESH KUMAR BHAGT  (6 years ago)

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

Is this solution Helpfull? Yes (4) | No (5)  

pankaj joshi  (6 years ago)

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

Is this solution Helpfull? Yes (4) | No   

  (6 years ago)

I think it should be 8c1+8c2+8c3+8c4+8c5+8c6+8c7+8c8=2^8-1=255

Is this solution Helpfull? Yes (3) | No (11)  

deeksha gupta  (6 years ago)

question from arun sharma example num:17.13

Is this solution Helpfull? Yes (3) | No (2)  

mansi sharma  (6 years ago)

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.

Is this solution Helpfull? Yes (1) | No (1)  

vikash  (5 years ago)

check out this link.It will help
https://gateoverflow.in/217975/%23permuatation?show=218002

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.