Combinatorics questions need help

Question

1)

You're going to a theme park with your family and they allow you to bring at least one but no more than $3$ friends with you.

If you choose from $8$ friends, how many different grouping can be formed?

2)

There are $5$ vegetable toppings and $3$ meat toppings available at a pizza shop.

How many different pizzas can you buy with $2$ different vegetable toppings and $1$ meat topping.

3)

Using your previous answer how many ways can you buy $4$ different pizzas.

Answer 1

1) Select between 1, 2, and 3 friends from 8: $\,^8C_1+\,^8C_2+\,^8C_3$

2) Select 2 of 5 vegetable toppings and 1 of 3 meat. $\,^5C_2\cdot\,^3C_1$

3) Select 4 from the result above. $\,^{(\,^5C_2\cdot\,^3C_1)}C_4$

Answer 2

I try to do it. $\color{red}{\text{(Corrections in red.)}}$

1.)you need to choose 1-3 fd from 8 fd.As the grroup do not need to have permutation , we use combination. That is $\,^8C_1+\,^8C_2+\,^8C_3=92 \color{red}{\checkmark}$

2.)I simply use symbol v=vegetable toppings and m=meat toppings To make a pizza, we should choose 2v and 1m,and there are 5v and 3m are available. we use combination again. thus,the number of different pizzas=$\,^5C_2 \color{red}{\not\!\!}+^{\color{red}{\ast}} \,^3C_1=\color{red}{\not\!}1\!\!\color{red}{\not\!}3\;\color{red}{30}$

$\color{red}{\text{By the universal principle of counting, you must multiply ways to perform separate tasks.}}$

AS there are $\color{red}{\not\!}1\!\!\color{red}{\not\!}3\;\color{red}{30}$ types of pizzas and we need to choose 4 diff. types from them thus, ways to buy 4 different pizzas=$\,^{\color{red}{\not}1\color{red}{\not}3\;\color{red}{30}}C_4=\color{red}{\not}7\!\!\!\color{red}{\not}1\!\!\!\color{red}{\not}5\; \color{red}{27405} \color{red}{\checkmark}$ Hope it can help you.:)