Question

Help

Answer 1

@Michele_Laino

Answer 2

Hold on I have to find it and attach it. Please wait.

Answer 3

ok!

Answer 4

U see it?

Answer 5

yes!

Answer 6

hint:
\[\left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right) = \frac{{6!}}{{3!3!}} = \frac{{3! \cdot 4 \cdot 5 \cdot 6}}{{3! \cdot 6}} = ...?\]

Answer 7

20?

Answer 8

yes!

Answer 9

Can you help with some more please?

Answer 10

ok!

Answer 11

I think the first option, since the set of three toppings can not be a subset of the set pizza

Answer 12

That's what I thought

Answer 13

That's the next two questions

Answer 14

brb

Answer 15

ok!

Answer 16

I'm back sorry my dad wanted me

Answer 17

ok!

Answer 18

Can you help me now?? @Michele_Laino

Answer 19

I'm ready!

Answer 20

Okay do you need me to re-post the question's?

Answer 21

post your question here, please

Answer 22

You posted it on here

Answer 23

I think that the number of ways, is equal to the number of subset of three elements from a set of 8 elements, and that number is:
\[\left( {\begin{array}{*{20}{c}}
8 \\
3
\end{array}} \right) = \frac{{8!}}{{3!3!}} = \frac{{5! \cdot 6 \cdot 7 \cdot 8}}{{3! \cdot 5!}} = ...?\]

Answer 24

oops...
\[\left( {\begin{array}{*{20}{c}}
8 \\
3
\end{array}} \right) = \frac{{8!}}{{3!5!}} = \frac{{5! \cdot 6 \cdot 7 \cdot 8}}{{3! \cdot 5!}} = ...?\]

Answer 25

So do it this way^^?

Answer 26

I got 112 but that is not a choice..

Answer 27

since in our computation, we can include the repeated subsets, which differ from the order of their elements only

Answer 28

Huh?

Answer 29

I got 56: