# Math Help - counting possibilities

1. ## counting possibilities

Can someone pleae help me urgently..... I feel really silly at the moment. My daughter came home with 2 maths problems for homework and I really dont know how to help her with these. can someone please explain how to work out the answers to the following problems so I can explain it to her.

1. you have 5 pizza toppings(sausage, onion, peppers, pepperoni and mushroom), how many different varities of pizzas can you make with these toppings?

2. 3 girls (Carry, Zoe and Lisa) go to the creek each has a backpack, they find 4 rocks, how many ways can the girls bring the rocks home?

2. Originally Posted by CasyC77
1. you have 5 pizza toppings(sausage, onion, peppers, pepperoni and mushroom), how many different varities of pizzas can you make with these toppings?
${5\choose 3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!\cdot 2!}$
2. 3 girls (Carry, Zoe and Lisa) go to the creek each has a backpack, they find 4 rocks, how many ways can the girls bring the rocks home?
${ {3+4-1}\choose 4 } ={6\choose 4} = \frac{6!}{4!(6-4)!}=\frac{6!}{4!\cdot 2!}$

3. Sorry but that doesnt make sense to me at all. I need to show my 9 year old daughter, and she needs to show the teacher how she worked it out. Sorry I am not good at maths at all.

4. Originally Posted by CasyC77
Sorry but that doesnt make sense to me at all. I need to show my 9 year old daughter, and she needs to show the teacher how she worked it out. Sorry I am not good at maths at all.
So I will assume you want me to do this without any formulas?

Let me do the first one.
Let us call the possible pizzas A,B,C,D,E instead to make it easier to type.

Then below are all the possible ways to put 3 together:
Code:
A,B,C
A,B,D
A,B,E
A,C,D
A,C,E
A,D,E
B,C,D
B,C,E
B,D,E
C,D,E
So there are a total of 10 different ways to do this.

5. We sat down on the weekend and worked out that there was more then 10 ways eg.
single toppings
Mushroom pizza
pepperoni pizza
onion pizza
pepper pizza
sausage pizza
then 2 toppings
mushroom and pepperoni
onion and pepper
etc etc etc
then 3 toppings
pepper sausage and onion
onion pepperoni and sausage
etc etc etc
then 4 toppings
pepper, sausage, onion and pepperoni
etc etc etc
then 5 toppings
pepper, sausage, onion, pepperoni and mushroom

does this make sense? So with this detail how many could you make?

6. Originally Posted by CasyC77

does this make sense? So with this detail how many could you make?
I am sorry, when you said the work "these" it looked like "three" to me. Thus, I assumed you were asking "how many can you make if you use 3 different one ... ". I have to leave now, maybe, tomorrow I will answer.

7. Originally Posted by CasyC77
Can someone pleae help me urgently..... I feel really silly at the moment. My daughter came home with 2 maths problems for homework and I really dont know how to help her with these. can someone please explain how to work out the answers to the following problems so I can explain it to her.

1. you have 5 pizza toppings(sausage, onion, peppers, pepperoni and mushroom),
how many different varities of pizzas can you make with these toppings?
There is 1 pizza with 0 toppings

There are 5 pizzas with 1 topping

There are 4 pizzas with sausage and one other topping, there are three with
onion but not sausage, there two pizzas with peppers and something else
other than sausage or onion, and one with peperoni and mushrooms, so
there are a total of 4+3+2+1=10 two topping pizzas.

Now there are as many three topping pizzas as there are two topping pizzas
as each pair of toppings leaves three toppings out, so there are 10 three
topping pizzas.

There are 5 four topping pizzas (each pizza leaves out one topping)

There is one five topping pizza.

So altogether we have:

1 + 5 + 10 + 10 + 5 +1 = 32

different pizzas.

RonL

8. Originally Posted by ThePerfectHacker
So I will assume you want me to do this without any formulas?

Let me do the first one.
Let us call the possible pizzas A,B,C,D,E instead to make it easier to type.

Then below are all the possible ways to put 3 together:
Code:
A,B,C
A,B,D
A,B,E
A,C,D
A,C,E
A,D,E
B,C,D
B,C,E
B,D,E
C,D,E
So there are a total of 10 different ways to do this.
are we supposed to only use 3 toppings at a time? i think we want the total number of combinations we can make with all five toppings. that is, how many kinds of pizzas we can make using one topping + how many kinds of pizza we can make using any two toppings + how many kinds of pizzas we can make using any 3 toppings + ...

the answer i think will amount to ${5 \choose 1} + {5 \choose 2} + {5 \choose 3} + {5 \choose 4} + {5 \choose 5}$

but since this is a 9 year old, let's not talk about combnations in the combinatoric sense.

we have sausage, onion, peppers, pepperoni and mushroom. as TPH did. let's just call them A, B, C, D and E for short.

How many ways can we make a pizza with one topping?

5 ways:
A
B
C
D
E

How many ways can we make a pizza with two different toppings?

10 ways:

AB
AC
AE
BC
BD
BE
CD
CE
DE

How many ways can we make a pizza with three different toppings?

10 ways:

ABC
ABD
ABE
ACD
ACE
BCD
BCE
BDE
CDE

How many ways can we make a pizza with four different toppings?

5 ways:

ABCD
ABCE
BCDA
BCDE
BDEA

How many ways can we make a pizza with five different toppings?

1 way:

ABCDE

So in all, there are 31 ways we can make a pizza with these toppings.

Did you see a pattern in the way i listed the different toppings in each category? If not, say so

EDIT: CaptainBlack saw fit to count the pizza with none of the toppings and got 32. i thought about that, but wasn't sure. Let's go with the Captain's answer, 32 ways

9. Thank you so much for you help on that question. I thought that was how to do it but I wasnt sure. Could you please help me with question 2 as well. Thanks again.

3 girls (Carry, Zoe and Lisa) go to the creek each has a backpack, they find 4 rocks, how many ways can the girls bring the rocks home?

10. Hello, CasyC77!

1. You have 5 pizza toppings (sausage, onion, peppers, pepperoni, mushroom).
How many different varieties of pizzas can you make with these toppings?

For each topping, you make a decision: Yes or No.

Look at the pattern that develops.

With Sausage, there are two choices: Yes or No.

For each of those two choices, there are two choices for Onion.
. . There are four choices regarding Sausage/Onion.

For each of those four choices, there two choices for Peppers.
. . There are eight choices regarding Sausage/Onion/Peppers.

We can see that, for each additional topping, the number of choices doubles.

Hence, for five toppings, there are: . $2 \times 2 \times 2 \times 2 \times 2 \:=\:32$ choices.

Therefore, there are 32 varieties of pizzas available
. . from "plain" (no toppings) to "the works" (all five toppings).

11. Hello, CasyC77!

Three girls (Carry, Zoe and Lisa) go to the creek; each has a backpack.
They find 4 rocks.. How many ways can the girls bring the rocks home?
I'll use the same approach for this problem.

For each rock, we must decide who to give it to: Carry, Zoe, or Lisa.
. . There are three choices.

For the first rock, there are three choices of carrier.

For each of those three choices, there are three choices for the second rock.
. . There are: $3 \times 3 \:=\:9$ choices.

For each of those nine choices, there are three choices for the third rock.
. . There are: $9 \times 3 \:=\:27$ choices.

For each of those twenty-seven choices, there are three choices for the fourth rock.
. . There are: $27 \times 3 \:=\:81$ choices.

Therefore, there are 81 ways the girls can bring the rocks home.

12. Originally Posted by Jhevon

EDIT: CaptainBlack saw fit to count the pizza with none of the toppings and got 32. i thought about that, but wasn't sure. Let's go with the Captain's answer, 32 ways
I eat a lot of pizza and since a Margarita has just cheese and sauce zero
toppings is indeed an option in the real world, what ever that is.

RonL

13. Originally Posted by Jhevon

the answer i think will amount to ${5 \choose 1} + {5 \choose 2} + {5 \choose 3} + {5 \choose 4} + {5 \choose 5}$
Yes, but you can use the trick:

${n\choose 1}+{n\choose 2}+...+{n\choose n} = 2^n-1$

14. Originally Posted by ThePerfectHacker
Yes, but you can use the trick:

${n\choose 1}+{n\choose 2}+...+{n\choose n} = 2^n-1$
yeah, i noticed that afterwards. and if we include ${n \choose 0}$ as CaptainBlack and the others suggest, we would get $2^n$, which vindicates Soroban's method