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?

#Combinatorics

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

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 :

$\space$

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

$\space$

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.

Arian Tashakkor - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Choosing nnn candies from mmm brands

Comments

Choosing $n$ candies from $m$ brands