- #1
vector
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Hello Everyone,
There is one interesting exercise in which it is asked to solve the following problem:
In how many ways can we distribute 20 candies among 6 children so that the youngest gets at most 2 candies?
This is my version of the solution:
Case 1: youngest child gets no candies, thus we have to distribute the 20 candies among 5 children, so we have C(20+5-1, 5) = 42 504
Case 2: youngest child gets one candy, thus we give him one candy, and then we have 19 candies to distribute between the other 5 children: C(19+5-1, 5) = 33 649
Case 3: youngest child gets two candies, so we give him two, and then we have 18 candies to distribute among the other 5 children: C(18+5-1, 5) = 26 334
And the total comes to 42 504 + 26 334 + 33 649 = 102 487 ways to distribute the candies.
This is the instructor's version of the solution, whose answer matches the answer in the study guide:
You have x1 + x2 + x3 + x4 + x5 + x6 = 20
Nw we will discuss x1, consider y1=x1-i where i is 0, 1, or 2..
So for i=0 we still have x1=y1 and then we still have x1 + x2 + x3 + x4 + x5 + x6 = 20 now I think we can consider x1 as x1=0 is a "choice".
To understand my point if you consider x1=1 then using you thinking we will get to x2 + x3 + x4 + x5 + x6 =19 and the answer will be C(19+5-1,19)=C(23,19) however.
If you consider i=1 so x1=y1+1 and then the answer will be C(19+6-1,16)=C(24,19), so 6 is always kept for all cases.
My question is: which solution is correct?
There is one interesting exercise in which it is asked to solve the following problem:
In how many ways can we distribute 20 candies among 6 children so that the youngest gets at most 2 candies?
This is my version of the solution:
Case 1: youngest child gets no candies, thus we have to distribute the 20 candies among 5 children, so we have C(20+5-1, 5) = 42 504
Case 2: youngest child gets one candy, thus we give him one candy, and then we have 19 candies to distribute between the other 5 children: C(19+5-1, 5) = 33 649
Case 3: youngest child gets two candies, so we give him two, and then we have 18 candies to distribute among the other 5 children: C(18+5-1, 5) = 26 334
And the total comes to 42 504 + 26 334 + 33 649 = 102 487 ways to distribute the candies.
This is the instructor's version of the solution, whose answer matches the answer in the study guide:
You have x1 + x2 + x3 + x4 + x5 + x6 = 20
Nw we will discuss x1, consider y1=x1-i where i is 0, 1, or 2..
So for i=0 we still have x1=y1 and then we still have x1 + x2 + x3 + x4 + x5 + x6 = 20 now I think we can consider x1 as x1=0 is a "choice".
To understand my point if you consider x1=1 then using you thinking we will get to x2 + x3 + x4 + x5 + x6 =19 and the answer will be C(19+5-1,19)=C(23,19) however.
If you consider i=1 so x1=y1+1 and then the answer will be C(19+6-1,16)=C(24,19), so 6 is always kept for all cases.
My question is: which solution is correct?