In how many ways can $4$ colas, $3$ iced teas, and $3$ orange juices be distributed to $10$ graduates if each grad is to receive $1$ beverage?

Question

In how many ways can four colas, three iced teas, and three orange juices be distributed to ten graduates if each grad is to receive one beverage?

$\binom{10}4\times\binom63\times1 = 4200$ ways

Is the answer and solution correct?

Answer 1

Yes. The answer and solution is correct.

Another way to do it is:

You have $10$ objects of which $4$ are colas, $3$ are iced teas and $3$ are orange juices. Basically, you have to permute all of them because out all $10$ need to be chosen

For permutation of $n$ objects of which $P$ are alike, $Q$ are alike and $R$ are alike, such that $P+Q+R=n$ is given by:

$$\frac{n}{P! Q! R!}$$

Hence, here $n=10, P=4, Q=3$ and $R=3.$

On substituting the values, we get $4200$ as the answer.