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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?
Correct answer is '16'. Can you explain this answer?
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In how many ways can 7 identical erasers be distributed among 4 kids ...
Distributing 7 identical erasers among 4 kids
To solve this problem, we can use the concept of stars and bars. The stars and bars method is a combinatorial technique used to count the number of ways to distribute identical objects among distinct containers.
Step 1: Distributing one eraser to each kid
Since each kid must receive at least one eraser, we start by giving one eraser to each of the 4 kids. After this initial distribution, we are left with 7 - 4 = 3 erasers.
Step 2: Distributing the remaining erasers using stars and bars
Now, we need to distribute the remaining 3 erasers among the 4 kids without any restrictions. We can represent the remaining erasers as stars and the spaces between the kids as bars.
For example, let's say we have 3 erasers represented by stars: ***
We can place the bars in different ways to represent different distributions:
- **|*|*|* (Kid 1 gets 2 erasers, Kid 2 gets 1 eraser, Kid 3 gets 1 eraser, Kid 4 gets 1 eraser)
- *|*|**|* (Kid 1 gets 1 eraser, Kid 2 gets 1 eraser, Kid 3 gets 2 erasers, Kid 4 gets 1 eraser)
- *|**|*|* (Kid 1 gets 1 eraser, Kid 2 gets 2 erasers, Kid 3 gets 1 eraser, Kid 4 gets 1 eraser)
- *|*|*|** (Kid 1 gets 1 eraser, Kid 2 gets 1 eraser, Kid 3 gets 1 eraser, Kid 4 gets 2 erasers)
Step 3: Counting the distributions
To count the number of distributions, we need to determine the number of ways we can arrange the stars and bars.
In this case, we have 3 stars and 3 + 1 = 4 bars. The stars can be placed in 6 possible positions (the 4 spaces between the bars and the 2 spaces on the ends).
Using the combination formula, we can calculate the number of ways to distribute the erasers as:
C(6,3) = 6! / (3! * (6-3)!) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20
Therefore, there are 20 different ways to distribute the remaining 3 erasers among the 4 kids.
Step 4: Calculating the total number of distributions
To calculate the total number of distributions, we multiply the number of ways to distribute the initial erasers (1 way) by the number of ways to distribute the remaining erasers (20 ways).
1 * 20 = 20
Therefore, there are 20 different ways to distribute 7 identical erasers among 4 kids, satisfying the given conditions.
To solve this problem, we can use the concept of stars and bars. The stars and bars method is a combinatorial technique used to count the number of ways to distribute identical objects among distinct containers.
Step 1: Distributing one eraser to each kid
Since each kid must receive at least one eraser, we start by giving one eraser to each of the 4 kids. After this initial distribution, we are left with 7 - 4 = 3 erasers.
Step 2: Distributing the remaining erasers using stars and bars
Now, we need to distribute the remaining 3 erasers among the 4 kids without any restrictions. We can represent the remaining erasers as stars and the spaces between the kids as bars.
For example, let's say we have 3 erasers represented by stars: ***
We can place the bars in different ways to represent different distributions:
- **|*|*|* (Kid 1 gets 2 erasers, Kid 2 gets 1 eraser, Kid 3 gets 1 eraser, Kid 4 gets 1 eraser)
- *|*|**|* (Kid 1 gets 1 eraser, Kid 2 gets 1 eraser, Kid 3 gets 2 erasers, Kid 4 gets 1 eraser)
- *|**|*|* (Kid 1 gets 1 eraser, Kid 2 gets 2 erasers, Kid 3 gets 1 eraser, Kid 4 gets 1 eraser)
- *|*|*|** (Kid 1 gets 1 eraser, Kid 2 gets 1 eraser, Kid 3 gets 1 eraser, Kid 4 gets 2 erasers)
Step 3: Counting the distributions
To count the number of distributions, we need to determine the number of ways we can arrange the stars and bars.
In this case, we have 3 stars and 3 + 1 = 4 bars. The stars can be placed in 6 possible positions (the 4 spaces between the bars and the 2 spaces on the ends).
Using the combination formula, we can calculate the number of ways to distribute the erasers as:
C(6,3) = 6! / (3! * (6-3)!) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20
Therefore, there are 20 different ways to distribute the remaining 3 erasers among the 4 kids.
Step 4: Calculating the total number of distributions
To calculate the total number of distributions, we multiply the number of ways to distribute the initial erasers (1 way) by the number of ways to distribute the remaining erasers (20 ways).
1 * 20 = 20
Therefore, there are 20 different ways to distribute 7 identical erasers among 4 kids, satisfying the given conditions.
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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
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 in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?Correct answer is '16'. Can you explain this answer?
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