SOLUTION: a boy have 5 coins each of different denominations. How many different sums of money he can form ? Its answer is 31 plz help
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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): You can put this solution on YOUR website!
Let's assume that every combination of coins has different value.
For instance, coin denominations can be powers of two, or three, etc.
There are combinations for coins,
Probably, the authors of the textbook don't consider 0 a valid sum.
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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