Permutation / Combinatorics Glue Method

[VadimC819]

Can an instruction please explain how to use the "Glue Method" / "Anagram Method" for the *Meats arrangement portion* in the question below?

A pizza shop can be created from any combination of 3 types of spice, 7 types of meat, among which are pepperoni and anchovies, and 4 types of cheese. If a customer wants to order a pizza with 1 type of spice, 2 types of cheese and 4 types of meat but without pepperoni and anchovies, how many possible ways to garnish the pizza are available to the pizza?

The answer is 450 to the entire question, but my question relates to how the 25 was obtained for the meat combination using Manhattan's "Glue method"

Thank you!

[esledge]

Hi Vadim, I can't find a reference to a "Glue Method" in any of our current books, and I couldn't find this question in our materials. If you could confirm the exact source (and exact wording) of this, I can give you a better answer.

For the question as you have written it, I actually get 90 as the final answer and 5 as the number of meat combos:

A pizza shop can be created from any combination of 3 types of spice, 7 types of meat, among which are pepperoni and anchovies, and 4 types of cheese. If a customer wants to order a pizza with 1 type of spice, 2 types of cheese and 4 types of meat but without pepperoni and anchovies, how many possible ways to garnish the pizza are available to the pizza?

The answer is 450 to the entire question, but my question relates to how the 25 was obtained for the meat combination using Manhattan's "Glue method"

Here's my thinking:
There are 3 ways to choose 1 spice from the 3 options (3 = 3!/1!2!).
There are 6 ways to choose 2 cheeses from the 4 options (6 = 4!/2!2! or using the slot method, 4*3 = 12 then divide by 2 since order doesn't matter).
If you are avoiding pepperoni and anchovies, then there are really only 7 - 2 = 5 types of meat to choose from.
There are 5 ways to choose 4 meats from the 5 allowed options (5!/4!1! or think of it as 5 ways to "leave out" one of the meats on the allowed list).

Overall, that's 3*6*5 = 90 pizzas. One thing that's not clear from the wording above is whether the cheeses or meats have to be "different types." For example, could the 2 cheese selections be mozzarella and extra mozzarella? My answer assumes that the answer is "no," but if it is "yes," that would increase the number of pizza combinations.