In how many ways can 8 identical chocolates be distributed among three children such that each can get any number of chocolates (including zero)?
68
24
45
21
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
Put n = 8, r = 3
Total number of ways =
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