If $10$ people have dinner together, how many
different ways can three(people) order chicken, four order steak and
three orders lobster?
So for this question would this be a permutation with similar objects or a partition with different items. If so I know one uses the formula.
$$\frac{n!}{n_{1}!n_{2}!n_{k}} = \frac{10!}{3!*4!*3!} =4200 \text{ ways }$$
Would this be the correct answer?
Best Answer
Imagine instead you have 10 different slots (1:10), 3 white balls, 4 green and 3 blue. In how many ways can you allocate these 10 balls into 10 bins? Obviously it is $\binom{10}{3}\binom{7}{4}\binom{3}{3}$, which is $\frac{10!}{3!4!3!}$