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In how many ways can 8 identical chocolates be distributed among three children such that each can get any number of chocolates (including zero)?

Option: 1

68


Option: 2

24


Option: 3

45


Option: 4

21


Answers (1)

best_answer

Let the first child gets 'a' number of chocolates, second gets ''b and third gets 'c'

So a + b + c = 8

And a, b, c can be whole numbers (as 0 chocolates may be given to a child) 

So number of ways of distribution = number of whole number solutions of the above equation, which is ^{n+r-1} C_{r-1}

Put n = 8, r = 3

Total number of ways = ^{8+3-1}C_{3-1}=45

Posted by

jitender.kumar

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