Question 364075: a boy have 5 coins each of different denominations. How many different sums of money he can form ? Its answer is 31 plz help
Found 4 solutions by Sphinx pinastri, sudhanshu_kmr, mzaffar241, greenestamps:
Answer by Sphinx pinastri(17)   (Show Source):
Answer by sudhanshu_kmr(1152)  (Show Source):
You can put this solution on YOUR website! As there are 5 coins, for every sum any particular coin may be selected or not.
thus, for each sum, there are two possibilities of each coin.
for example, let A,B,C,D AND E are coins, then sum can be get
ABCDE
1 1 1 1 1
1 1 1 1 0
:
:
:
0 0 0 0 1
0 0 0 0 0 ( no one coin selected)
Here 1 represent it is included and 0 represent not included.
total no. of different sum = 2^5 - 1 = 31
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Best of luck.......
Answer by mzaffar241(1)  (Show Source):
You can put this solution on YOUR website! Selecting 1 coin out of 5 coins: 5C1=5 different sums
Selecting 2 coins out of 4 coins: 5C2=10 different sums
Selecting 3 coins out of 4 coins: 5C2=10 different sums
Selection 4 coins out of 4 coins: 5C4=5 different sums
Selection 5 coins out of 4 coins 5C5=1 different sums
Hence total number of different sums of money :5+10+10+5+1= 31 Answer
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
In making a sum of money, he has 2 choices for each coin: include it, or not. With 5 coins, the number of combinations of choices he can make -- and therefore the number of different sums of money he can make -- is 2*2*2*2*2 = 2^5 = 32.
One of those possibilities is 0 (if he chooses not to include every coin in his selection). Since you say the answer to the question is 31, apparently a sum of 0 is not allowed.
Since all the other sums are allowed, the number of (nonzero) sums he can make is 32-1 = 31.
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