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?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestIf \(n\) 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 :

\(\space\)

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

\(\space\)

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.

Log in to reply