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The number of ways in which 6 different balls can be put in two boxes of different sizes so that no box remains empty is
- a)62
- b)64
- c)36
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The number of ways in which 6 different balls can be put in two boxes ...
Each ball can be put in 2 ways (either in one box or the other)
∴ 6 balls can be put in 2 x 2 x ... to six times, i.e.,26 ways. But in two of the ways one box is empty. So, the required number of ways = 26-2
∴ 6 balls can be put in 2 x 2 x ... to six times, i.e.,26 ways. But in two of the ways one box is empty. So, the required number of ways = 26-2
Most Upvoted Answer
The number of ways in which 6 different balls can be put in two boxes ...
To solve this problem, we can use the concept of distributing objects into boxes.
First, let's consider the two boxes as Box 1 and Box 2.
Case 1: Box 1 has 1 ball
In this case, we can choose any one of the 6 balls to be put in Box 1. Once we choose a ball for Box 1, the remaining 5 balls can be put in Box 2 in different ways. Therefore, the number of ways to distribute the balls in this case is 6 * 5 = 30.
Case 2: Box 1 has 2 balls
In this case, we need to choose 2 balls from the 6 available balls to put in Box 1. This can be done in 6C2 ways (6 choose 2), which is equal to 15. Once we choose the 2 balls for Box 1, the remaining 4 balls can be put in Box 2 in different ways. Therefore, the number of ways to distribute the balls in this case is 15 * 4 = 60.
Case 3: Box 1 has 3 balls
In this case, we need to choose 3 balls from the 6 available balls to put in Box 1. This can be done in 6C3 ways, which is equal to 20. Once we choose the 3 balls for Box 1, the remaining 3 balls can be put in Box 2 in different ways. Therefore, the number of ways to distribute the balls in this case is 20 * 3 = 60.
Case 4: Box 1 has 4 balls
In this case, we need to choose 4 balls from the 6 available balls to put in Box 1. This can be done in 6C4 ways, which is equal to 15. Once we choose the 4 balls for Box 1, the remaining 2 balls can be put in Box 2 in different ways. Therefore, the number of ways to distribute the balls in this case is 15 * 2 = 30.
Case 5: Box 1 has 5 balls
In this case, we need to choose 5 balls from the 6 available balls to put in Box 1. This can be done in 6C5 ways, which is equal to 6. Once we choose the 5 balls for Box 1, the remaining 1 ball can be put in Box 2 in only 1 way. Therefore, the number of ways to distribute the balls in this case is 6 * 1 = 6.
Case 6: Box 1 has 6 balls
In this case, all 6 balls must be put in Box 1. There is no way to distribute the balls in any other way.
Finally, to get the total number of ways, we add up the number of ways from each case:
30 + 60 + 60 + 30 + 6 + 1 = 187
However, we need to divide this number by 2 since the two boxes are of different sizes (one is larger and one is smaller). This is because switching the roles of Box 1 and Box 2 would result in the same distribution. Therefore, the final number of ways is 187/2 = 93.5
Since the number of ways must be a
First, let's consider the two boxes as Box 1 and Box 2.
Case 1: Box 1 has 1 ball
In this case, we can choose any one of the 6 balls to be put in Box 1. Once we choose a ball for Box 1, the remaining 5 balls can be put in Box 2 in different ways. Therefore, the number of ways to distribute the balls in this case is 6 * 5 = 30.
Case 2: Box 1 has 2 balls
In this case, we need to choose 2 balls from the 6 available balls to put in Box 1. This can be done in 6C2 ways (6 choose 2), which is equal to 15. Once we choose the 2 balls for Box 1, the remaining 4 balls can be put in Box 2 in different ways. Therefore, the number of ways to distribute the balls in this case is 15 * 4 = 60.
Case 3: Box 1 has 3 balls
In this case, we need to choose 3 balls from the 6 available balls to put in Box 1. This can be done in 6C3 ways, which is equal to 20. Once we choose the 3 balls for Box 1, the remaining 3 balls can be put in Box 2 in different ways. Therefore, the number of ways to distribute the balls in this case is 20 * 3 = 60.
Case 4: Box 1 has 4 balls
In this case, we need to choose 4 balls from the 6 available balls to put in Box 1. This can be done in 6C4 ways, which is equal to 15. Once we choose the 4 balls for Box 1, the remaining 2 balls can be put in Box 2 in different ways. Therefore, the number of ways to distribute the balls in this case is 15 * 2 = 30.
Case 5: Box 1 has 5 balls
In this case, we need to choose 5 balls from the 6 available balls to put in Box 1. This can be done in 6C5 ways, which is equal to 6. Once we choose the 5 balls for Box 1, the remaining 1 ball can be put in Box 2 in only 1 way. Therefore, the number of ways to distribute the balls in this case is 6 * 1 = 6.
Case 6: Box 1 has 6 balls
In this case, all 6 balls must be put in Box 1. There is no way to distribute the balls in any other way.
Finally, to get the total number of ways, we add up the number of ways from each case:
30 + 60 + 60 + 30 + 6 + 1 = 187
However, we need to divide this number by 2 since the two boxes are of different sizes (one is larger and one is smaller). This is because switching the roles of Box 1 and Box 2 would result in the same distribution. Therefore, the final number of ways is 187/2 = 93.5
Since the number of ways must be a
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Community Answer
The number of ways in which 6 different balls can be put in two boxes ...
Each ball can be put in 2 ways (either in one box or the other)
∴ 6 balls can be put in 2 x 2 x ... to six times, i.e.,26 ways. But in two of the ways one box is empty. So, the required number of ways = 26-2
∴ 6 balls can be put in 2 x 2 x ... to six times, i.e.,26 ways. But in two of the ways one box is empty. So, the required number of ways = 26-2
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The number of ways in which 6 different balls can be put in two boxes of different sizes so that no box remains empty isa)62b)64c)36d)none of theseCorrect answer is option 'A'. Can you explain this answer?
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