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Clarification in a problem

Hey Guys, I have this question but my friend and I are getting different answers, and I don't have the answer booklet with me. Could you guys help ?

So, here's the question.

Find the number of all possible k-tuples of non - negative integers (n(1),n(2),n(3).... n(k)) such that

sum_{i=1}^k n(i) = 100

So, this is the question. Now here's my solution.

Let there be 100 stars placed in a line. We have to divide these 100 stars into k groups such that no group equals 0 . So, let me do the division between stars by placing a bar between the stars. By this way, there are 99 possible ways in which I can set a bar (between any two stars) . But as I only have to create k groups, I need only place k-1 bars..

So I have 99 places and I need to choose any (k-1) places.. SO, I get the answer as

99 C (k-1)

Is it correct guys ? Because I have a friend who says the answer is (99+k)C(k-1).

Note by Adeetya Tantia
2 years, 6 months ago

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Your friend is correct.

If the problem asked about positive integers, then your answer of \(\dbinom{99}{k - 1}\) would be correct. But the \(n_i\) are nonnegative, which means they can be 0. Your approach does not account for this.

Instead, consider any arrangement of 100 stars and \(k - 1\) bars. (In particular, there could be two or more bars between consecutive stars.) Then you take \(n_1\) as the number of stars before the first bar, \(n_2\) as the number of stars between the first bar and second bar, and so on. This gives you solutions where the \(n_i\) can be 0. Hence, the number of solutions is the number of ways of arranging 100 stars and \(k - 1\) bars, which is \[\binom{100 + (k - 1)}{100} = \binom{k + 99}{100} = \binom{k + 99}{k - 1}.\] Jon Haussmann · 2 years, 6 months ago

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@Jon Haussmann Thanks. Adeetya Tantia · 2 years, 6 months ago

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Use the Stars and Bars formula, it's the same thing. A standard result ! Aditya Raut · 2 years, 3 months ago

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