Question

If 8 people eat dinner together, in how many ways can 3 order chicken, 4 order steak and 1 order lobster (assuming the similar orders are identical, except for which person they are given to)

Answer 1

Solution:

Given:

$\begin{gathered} 8\text{ people eat dinner together} \\ 3\text{ order chicken} \\ 4\text{ order steak} \\ 1\text{ order lobster} \end{gathered}$

Number of ways to order chicken

$\begin{gathered} ^8C_3=\frac{8!}{(8-3)!3!} \\ =\frac{8!}{5!3!} \\ =\frac{8\times7\times6\times5!}{5!\times3\times2\times1} \\ =56\text{ ways} \end{gathered}$

Number of ways to order steak

$\begin{gathered} ^8C_4=\frac{8!}{(8-4)!4!} \\ =\frac{8!}{4!4!} \\ =\frac{8\times7\times6\times5\times4!}{4!\times4\times3\times2\times1} \\ =70\text{ ways} \end{gathered}$

Number of ways to order lobster

$\begin{gathered} ^8C_1=\frac{8!}{(8-1)!1!} \\ =\frac{8!}{7!1!} \\ =\frac{8\times7!}{7!\times1} \\ =8\text{ ways} \end{gathered}$

Hence, the number of ways one can order 3 chicken, 4 steaks, and 1 lobster is;

$56\times70\times8=31,360\text{ways}$

Therefore, the number of ways one can order 3 chicken, 4 steaks and 1 lobster is 31,360 ways