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A man has 3 sons and 6 schools within his reach. In how many ways, he can send them to school, if no two of his sons are to read in the same school?
- a)6P2
- b)6P3
- c)63
- d)36
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A man has 3 sons and 6 schools within his reach. In how many ways, he ...
The problem is asking us to find the number of ways to distribute 3 distinct objects (sons) into 6 distinct groups (schools) without repetition. This is a problem of permutations.
Total number of ways to distribute 3 sons into 6 schools without any restriction is 6^3 (since each son has 6 choices of school).
But since no two sons can go to the same school, we need to remove the cases where two or more sons go to the same school.
Number of ways to choose 2 schools out of 6 for 2 sons to go to is 6C2.
Number of ways to choose which 2 sons go to those schools is 3C2 (since we choose 2 out of 3 sons).
Number of ways to assign the remaining son to any of the remaining 4 schools is 4.
Therefore, number of ways to distribute the sons such that no two of them are in the same school is:
6^3 - 6C2 * 3C2 * 4
= 216 - 90
= 126
This is the same as 6P3, which is the answer in option (b).
Total number of ways to distribute 3 sons into 6 schools without any restriction is 6^3 (since each son has 6 choices of school).
But since no two sons can go to the same school, we need to remove the cases where two or more sons go to the same school.
Number of ways to choose 2 schools out of 6 for 2 sons to go to is 6C2.
Number of ways to choose which 2 sons go to those schools is 3C2 (since we choose 2 out of 3 sons).
Number of ways to assign the remaining son to any of the remaining 4 schools is 4.
Therefore, number of ways to distribute the sons such that no two of them are in the same school is:
6^3 - 6C2 * 3C2 * 4
= 216 - 90
= 126
This is the same as 6P3, which is the answer in option (b).
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Community Answer
A man has 3 sons and 6 schools within his reach. In how many ways, he ...
Given:
A man has 3 sons and he can choose from 6 schools.
To find:
In how many ways he can send them to school if no two sons are to read in the same school.
Solution:
Since no two sons can be in the same school, we can say that the first son can be sent to any of the 6 schools. The second son can be sent to any of the remaining 5 schools. Similarly, the third son can be sent to any of the remaining 4 schools.
Using the multiplication principle, the total number of ways in which the man can send his sons to school is:
6 x 5 x 4 = 120
However, we have overcounted as the order in which we send the sons to school does not matter. Hence, we need to divide by the number of ways in which we can arrange the 3 sons.
Using the permutation formula, the number of ways in which we can arrange 3 distinct objects is:
3P3 = 3!
= 3 x 2 x 1
= 6
Therefore, the total number of ways in which the man can send his sons to school if no two sons are to read in the same school is:
120/6 = 20
Hence, the correct option is (b) 6P3, which is equal to 6 x 5 x 4.
A man has 3 sons and he can choose from 6 schools.
To find:
In how many ways he can send them to school if no two sons are to read in the same school.
Solution:
Since no two sons can be in the same school, we can say that the first son can be sent to any of the 6 schools. The second son can be sent to any of the remaining 5 schools. Similarly, the third son can be sent to any of the remaining 4 schools.
Using the multiplication principle, the total number of ways in which the man can send his sons to school is:
6 x 5 x 4 = 120
However, we have overcounted as the order in which we send the sons to school does not matter. Hence, we need to divide by the number of ways in which we can arrange the 3 sons.
Using the permutation formula, the number of ways in which we can arrange 3 distinct objects is:
3P3 = 3!
= 3 x 2 x 1
= 6
Therefore, the total number of ways in which the man can send his sons to school if no two sons are to read in the same school is:
120/6 = 20
Hence, the correct option is (b) 6P3, which is equal to 6 x 5 x 4.
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A man has 3 sons and 6 schools within his reach. In how many ways, he can send them to school, if no two of his sons are to read in the same school?a)6P2b)6P3c)63d)36Correct answer is option 'B'. Can you explain this answer?
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