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a) In how many ways can nicole distribute 20 choclates among
A,B,C and D if each receives at least one?
b)If nicole may choose to give no choclate to any number of people whom she dislikes on that particular day?
a) My answer (20-4)C(4). Since everyone receives at least 1 choclate so the total no. of choclates reduces to 16. Is the logic correct???whereas the OA is 19C3.
b) No clues on the second one
A,B,C and D if each receives at least one?
b)If nicole may choose to give no choclate to any number of people whom she dislikes on that particular day?
a) My answer (20-4)C(4). Since everyone receives at least 1 choclate so the total no. of choclates reduces to 16. Is the logic correct???whereas the OA is 19C3.
b) No clues on the second one
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10 Nov 2006, 09:46
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can't follow your logic, but can explain the OA (and the correct one)
put the 20 chocolates on a line.
now put 3 "markers" between the chocolates. (not two markers are in the same place). this partition the line of chocolate into 4 segments. each segment contain at least one chocolate.
so the number of ways to distribute 20 chocolate to 4 distinct people is the same a the number of ways to partition a line of 20 items into 4 distinct segments.
now, you have 19 possible places to put these 3 markers (that is - between any 2 neighboring chocolates).
there are 19C3 ways of placing 3 (identical) markers in 19 places without repetition (no two markers are at the same place).
hope it clarifies the answer.
the answer to question b should be similar, but this time people are allowed to get 0. or in our interpretation, markers can be put in the ame place, and there are now 21 place to put the markers instead of 19 (because marker can be put before the first, meaning that peron A gets nothing, and after the last chocolate meaning that person D gets nothing.
now you can do the math on your own.
enjoy...
amit.
put the 20 chocolates on a line.
now put 3 "markers" between the chocolates. (not two markers are in the same place). this partition the line of chocolate into 4 segments. each segment contain at least one chocolate.
so the number of ways to distribute 20 chocolate to 4 distinct people is the same a the number of ways to partition a line of 20 items into 4 distinct segments.
now, you have 19 possible places to put these 3 markers (that is - between any 2 neighboring chocolates).
there are 19C3 ways of placing 3 (identical) markers in 19 places without repetition (no two markers are at the same place).
hope it clarifies the answer.
the answer to question b should be similar, but this time people are allowed to get 0. or in our interpretation, markers can be put in the ame place, and there are now 21 place to put the markers instead of 19 (because marker can be put before the first, meaning that peron A gets nothing, and after the last chocolate meaning that person D gets nothing.
now you can do the math on your own.
enjoy...
amit.
Thanks for explaining the 1st part.
the answer for the second part would be 21C3. rite?
the answer for the second part would be 21C3. rite?
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
