Question

At a local pizza parlor, patrons have 3 choices of cheese and 5 choices of meat. In how many different ways can a patron choose 1 type of cheese and 1 type of meat?

Answer 1

Answer:15

Total choices=meat choices×cheese choices
=(5/1)×(3/1)=5×3=15

Answer 2

At a local pizza parlor, patrons have 3 choices of cheese and 5 choices of meat. The number of different ways the patron can choose 1 type of cheese and 1 type of meat is 15 ways.

The combination of an object is the method used in selecting the possible number of arrangements in an array of data. It is given by the formula:

$\mathbf{^nC_r = \dfrac{n!}{r!(n-r)!}}$

where;

• $\mathbf{^nC_r =}$ number of combinations

• n = total no. of objects in the data set

• r = no of objects to be chosen in the data set.

From the given information:

• total number of data n = 5 and 3
• and the no of objects to be chosen r = 1 and 1

∴

we need to find the combination of the dataset to determine the ways in which the patron can choose 1 type of cheese and meat.

i.e.

$\mathbf{^nC_r = \dfrac{5!}{1!(5-1)!} \times \dfrac{3!}{1!(3-1)!}}$

$\mathbf{^nC_r = \dfrac{5!}{1!(4)!} \times \dfrac{3!}{1!(2)!}}$

$\mathbf{^nC_r = \dfrac{5 \times 4!}{1!(4)!} \times \dfrac{3 \times 2!}{1!(2)!}}$

$\mathbf{^nC_r = 5 \times 3}$

$\mathbf{^nC_r = 15 }$

Therefore, we can conclude that the number of ways that the patron can choose 1 type of cheese and 1 type of meat is 15 ways.

Learn more about combination here:

https://brainly.com/question/19692242?referrer=searchResults