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Choosing \(n\) candies from \(m\) brands

In how many ways can we choose \(n\) candies from \(m\) brands?

Note: Repeated selection from the same brand is allowed and \(n\leq m\).

Why is \(m^n\) not the correct answer?

Note by D K
1 year, 1 month ago

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If \(n\) candies are chosen from \(m\) brands, then the sum of the number of candies from each brand equals \(n\).(That's pretty obvious right?!).Suppose \(x_1\) candies are chosen from Brand #1 , \(x_2\) candies from Brand #2 and ... \(x_m\) candies from Brand #\(m\).Now continuing my argument above , obviously the answer to your question is equivalent to the number of answers to the equation :


\(x_1 + x_2 + \dots + x_m = m \) , \(x_i \ge 0\) ;


Now if you're familiar with "Stars and bars" you'd know that the answer is \( n+m-1 \choose m-1 \).(If not , you can read it's wikipage here , I'm too lazy to write the whole thing down here)

Now as for the answer to your second question,the answer \(m^n\) would definitely not be correct since you're not counting the cases where no candies are chosen from a particular brand also you're not talking into account the fact that candies from different brands are not alike.I hope I could help you get a good grasp on this. Arian Tashakkor · 1 year, 1 month ago

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