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

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=(8!)/((8-3)!3!) \\ =(8!)/(5!3!) \\ =(8*7*6*5!)/(5!*3*2*1) \\ =56\text{ ways} \end{gathered}`

Number of ways to order steak

`\begin{gathered} ^8C_4=(8!)/((8-4)!4!) \\ =(8!)/(4!4!) \\ =(8*7*6*5*4!)/(4!*4*3*2*1) \\ =70\text{ ways} \end{gathered}`

Number of ways to order lobster

`\begin{gathered} ^8C_1=(8!)/((8-1)!1!) \\ =(8!)/(7!1!) \\ =(8*7!)/(7!*1) \\ =8\text{ ways} \end{gathered}`

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

`56*70*8=31,360\text{ways}`

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