Chapter 10, Problem 3CA

To determine

To calculate:

The number of different fruit smoothies that can be ordered made from 2 liquid and 3 fruit ingredients.

Answer to Problem 3CA

The different number of fruit smoothies that can be ordered made from 2 liquid and 3 fruit ingredients are $1800$ .

Explanation of Solution

Given information:

  

Liquids and Fruits menu

Formula used:

The expression for a combination of $n$ objects taken $r$ at a time can be given as follows:

  $C{}_r=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

The number of possible outcomes as the number of combinations of $6$ ingredients of liquids taken $2$ at a time i.e. $C{}_2$ because the order is not significant.

Fruits can be chosen in the same way i.e. $C{}_3$ .

Use the combination of both fruits and liquid ingredients i.e. $C{}_2\times C{}_3$ .

Thus, the expression for a combination of $n$ objects taken $r$ at a time can be given as follows:

  $C{}_r=\frac{n!}{r!\left(n-r\right)!}$

Substitute the value of the $C{}_r$ i.e. $C{}_2$ and $C{}_3$ in above equation,

Where,

For $C{}_2$ ,

  $\begin{array}{l}n=6\\ r=2\end{array}$

For $C{}_3$ ,

  $\begin{array}{l}n=10\\ r=3\end{array}$

Thus, the required number of smoothies are:

  $\begin{array}{l}C{}_2\times C{}_3=\frac{6!}{2!\left(6-2\right)!}\times \frac{10!}{3!\left(10-3\right)!}\\ C{}_2\times C{}_3=\frac{6!}{2!\left(4\right)!}\times \frac{10!}{3!\left(7\right)!}\\ C{}_2\times C{}_3=\frac{6\times 5\times 4!}{2\times 1\times 4!}\times \frac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}\\ C{}_2\times C{}_3=\frac{6\times 5}{2\times 1}\times \frac{10\times 9\times 8}{3\times 2\times 1}\\ C{}_2\times C{}_3=\frac{30}{2}\times \frac{720}{6}\\ C{}_2\times C{}_3=15\times 120\\ C{}_2\times C{}_3=1800\end{array}$

Hence, the different number of fruit smoothies that can be ordered made from 2 liquid and 3 fruit ingredients are $1800$ .