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In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?
- a)16
- b)20
- c)14
- d)15
Correct answer is option 'A'. Can you explain this answer?
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In how many ways can 7 identical erasers be distributed among 4 kids ...
We have been given that a + b + c + d = 7
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Total ways of distributing 7 things among 4 people so that each one gets at least one = n-1Cr-1 = 6C3 = 20 Now we need to subtract the cases where any one person got more than 3 erasers. Any person cannot get more than 4 erasers since each child has to get at least 1. Any of the 4 childs can get 4 erasers. Thus, there are 4 cases. On subtracting these cases from the total cases we get the required answer. Hence, the required value is 20 - 4 = 16
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In how many ways can 7 identical erasers be distributed among 4 kids ...
To solve this problem, we can use the concept of distributing identical objects among distinct containers with restrictions. In this case, we need to distribute 7 identical erasers among 4 kids, ensuring that each kid gets at least one eraser and no kid gets more than 3 erasers.
Let's consider the possible scenarios:
1. One kid gets 3 erasers and the other three kids get 1 eraser each:
- There are 4 ways to choose which kid gets 3 erasers.
- There are 7 ways to choose 3 erasers out of the 7 identical erasers.
- There are 3! ways to distribute the remaining 4 erasers among the remaining 3 kids (as each kid gets 1 eraser).
- So, the total number of ways for this scenario is 4 * 7 * 3! = 84.
2. One kid gets 2 erasers, another kid gets 2 erasers, and the other two kids get 1 eraser each:
- There are 4C2 = 6 ways to choose which two kids get 2 erasers.
- There are 7 ways to choose 2 erasers out of the 7 identical erasers for the first kid.
- There are 5 ways to choose 2 erasers out of the remaining 5 erasers for the second kid.
- There are 2! ways to distribute the remaining 3 erasers among the remaining 2 kids (as each kid gets 1 eraser).
- So, the total number of ways for this scenario is 6 * 7 * 5 * 2! = 420.
3. One kid gets 2 erasers and the other three kids get 1 eraser each:
- There are 4 ways to choose which kid gets 2 erasers.
- There are 7 ways to choose 2 erasers out of the 7 identical erasers.
- There are 3! ways to distribute the remaining 5 erasers among the remaining 3 kids (as each kid gets 1 eraser).
- So, the total number of ways for this scenario is 4 * 7 * 3! = 168.
Adding up the total number of ways from all scenarios, we get 84 + 420 + 168 = 672 ways. However, this count includes scenarios where one or more kids get more than 3 erasers, which is not allowed.
Subtracting the invalid scenarios, we find that there are 672 - 4 = 668 ways to distribute the erasers. But this count also includes the scenario where all 7 erasers are given to one kid, which violates the condition that each kid should get at least one eraser.
So, the correct answer is the total count minus the invalid scenarios, which is 668 - 1 = 667 ways.
Thus, option 'A' (16) is not the correct answer.
Let's consider the possible scenarios:
1. One kid gets 3 erasers and the other three kids get 1 eraser each:
- There are 4 ways to choose which kid gets 3 erasers.
- There are 7 ways to choose 3 erasers out of the 7 identical erasers.
- There are 3! ways to distribute the remaining 4 erasers among the remaining 3 kids (as each kid gets 1 eraser).
- So, the total number of ways for this scenario is 4 * 7 * 3! = 84.
2. One kid gets 2 erasers, another kid gets 2 erasers, and the other two kids get 1 eraser each:
- There are 4C2 = 6 ways to choose which two kids get 2 erasers.
- There are 7 ways to choose 2 erasers out of the 7 identical erasers for the first kid.
- There are 5 ways to choose 2 erasers out of the remaining 5 erasers for the second kid.
- There are 2! ways to distribute the remaining 3 erasers among the remaining 2 kids (as each kid gets 1 eraser).
- So, the total number of ways for this scenario is 6 * 7 * 5 * 2! = 420.
3. One kid gets 2 erasers and the other three kids get 1 eraser each:
- There are 4 ways to choose which kid gets 2 erasers.
- There are 7 ways to choose 2 erasers out of the 7 identical erasers.
- There are 3! ways to distribute the remaining 5 erasers among the remaining 3 kids (as each kid gets 1 eraser).
- So, the total number of ways for this scenario is 4 * 7 * 3! = 168.
Adding up the total number of ways from all scenarios, we get 84 + 420 + 168 = 672 ways. However, this count includes scenarios where one or more kids get more than 3 erasers, which is not allowed.
Subtracting the invalid scenarios, we find that there are 672 - 4 = 668 ways to distribute the erasers. But this count also includes the scenario where all 7 erasers are given to one kid, which violates the condition that each kid should get at least one eraser.
So, the correct answer is the total count minus the invalid scenarios, which is 668 - 1 = 667 ways.
Thus, option 'A' (16) is not the correct answer.
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In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?a) 16b) 20c) 14d) 15Correct answer is option 'A'. Can you explain this answer?
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