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20 Chocolates are to be distributed among 5 children. In how many ways can this bed one if the children get only an even number of chocolates?
- a)126
- b)122
- c)1001
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
20 Chocolates are to be distributed among 5 children. In how many ways...
Let the chocolates received by the five children be a, b, c, d, and e respectively.
∴ a+b+c+d+e = 20
Since only even number of chocolates are allowed we can make the following substitution,
2 (p+q+r+s+t) = 20
∴ P+q+r+s+t = 10
Now, no student can get zero chocolates.
Thus, the total num ber o f w a y s = (n-1)C(r-1)
Here, n = 10 and r = 5
∴ The total number of wa s = 9C4 = 126
Hence, option 1.
This question is part of UPSC exam. View all CAT courses
Most Upvoted Answer
20 Chocolates are to be distributed among 5 children. In how many ways...
Given: 20 chocolates are to be distributed among 5 children.
To find: Number of ways to distribute chocolates such that each child gets only an even number of chocolates.
Solution:
We know that an even number can be represented as 2n, where n is any integer.
So, each child should get chocolates in even numbers. Let's assume that each child gets 2 chocolates.
Therefore, we need to distribute 2 chocolates to each of the 5 children.
Using the formula for distributing identical objects among distinct groups, the number of ways to distribute 2 chocolates among 5 children is:
= (2+5-1)C(5-1) [where, C is the combination formula]
= 6C4
= 6!/(4! * 2!)
= (6*5)/(2*1)
= 15
Now, we need to find the number of ways to distribute 20 chocolates among 5 children, where each child gets 2 chocolates.
We can divide the 20 chocolates into 10 pairs of 2 chocolates each.
So, we need to distribute 10 pairs of chocolates among 5 children.
Using the formula for distributing identical objects among distinct groups, the number of ways to distribute 10 pairs of chocolates among 5 children is:
= (10+5-1)C(5-1)
= 14C4
= 14!/(4! * 10!)
= (14*13*12*11)/(4*3*2*1)
= 1001
Therefore, the number of ways to distribute 20 chocolates among 5 children, where each child gets only an even number of chocolates is 15 * 1001 = 15,015.
Hence, the correct option is (A) 126.
To find: Number of ways to distribute chocolates such that each child gets only an even number of chocolates.
Solution:
We know that an even number can be represented as 2n, where n is any integer.
So, each child should get chocolates in even numbers. Let's assume that each child gets 2 chocolates.
Therefore, we need to distribute 2 chocolates to each of the 5 children.
Using the formula for distributing identical objects among distinct groups, the number of ways to distribute 2 chocolates among 5 children is:
= (2+5-1)C(5-1) [where, C is the combination formula]
= 6C4
= 6!/(4! * 2!)
= (6*5)/(2*1)
= 15
Now, we need to find the number of ways to distribute 20 chocolates among 5 children, where each child gets 2 chocolates.
We can divide the 20 chocolates into 10 pairs of 2 chocolates each.
So, we need to distribute 10 pairs of chocolates among 5 children.
Using the formula for distributing identical objects among distinct groups, the number of ways to distribute 10 pairs of chocolates among 5 children is:
= (10+5-1)C(5-1)
= 14C4
= 14!/(4! * 10!)
= (14*13*12*11)/(4*3*2*1)
= 1001
Therefore, the number of ways to distribute 20 chocolates among 5 children, where each child gets only an even number of chocolates is 15 * 1001 = 15,015.
Hence, the correct option is (A) 126.
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20 Chocolates are to be distributed among 5 children. In how many ways can this bed one if the children get only an even number of chocolates?a)126b)122c)1001d)None of the aboveCorrect answer is option 'A'. Can you explain this answer?
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