A person has 4 coins each of different denomination. What is the number of different sums of money the person can form (using one or more coins at a time) ?

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Q: 65 (IAS/2009)
A person has 4 coins each of different denomination. What is the number of different sums of money the person can form (using one or more coins at a time) ?

question_subject: 

Maths

question_exam: 

IAS

stats: 

0,8,3,3,8,0,0

keywords: 

{'coins': [2, 0, 1, 1], 'more coins': [0, 0, 1, 0], 'different sums': [0, 0, 2, 0], 'different denomination': [0, 0, 1, 0], 'number': [0, 0, 0, 2], 'person': [3, 1, 18, 13]}

The question asks about the total number of different sums a person can form using four coins of different denominations. To solve this, one needs to understand the principle of combinations in probability. The situation represents a power set, a set of all subsets of a set, including both the empty set and the set itself.

Each coin can either be included (1) or not included (0) in each sum. So, for 4 coins, 2^4 possibilities exist which equals 16. But one of these possibilities is the case where no coin is used, which means no sum is achieved. So that possibility should be eliminated.

Now, let`s break down the answer options:

Option 1: 16 - This is the total number of subsets including no sum, so it`s not correct for this question.

Option 2: 15 - This is the correct answer, it represents the combinations of the 4 different coins excluding the case with no sum.

Option 3: 12 - This number is less than the correct number of combinations, so this option is incorrect.

Option 4: 11 - This number is also less than the correct number of combinations, which makes this option incorrect as well.

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