How did they solved this problem i got this one wrong on my quizz.

Pizza Varieties A television commercial for Little Caesars
pizza announced that with the purchase of two pizzas, one could receive free any combination of up to five toppings
on each pizza. The commercial shows a young child waiting in line at Little Caesars who calculates that there are
1,048,576 possibilities for the toppings on the two pizzas.*

a. Verify the child’s calculation. Use the fact that Little Caesars has 11 toppings to choose from. Assume that the order of the two pizzas matters; that is, if the first pizza
has combination 1 and the second pizza has combination
2, that is different from combination 2 on the first pizza and combination 1 on the second.

The answer that's in the back of the book: 1, 048,574

b. In a letter to The Mathematics Teacher, Joseph F. Heiser
argued that the two combinations described in part a
should be counted as the same, so the child has actually
overcounted. Give the number of possibilities if the
order of the two pizzas doesn’t matter.

The answer they got was: 524,800

Consider first the number of ways a pizza can be prepared with 0, 1, 2, 3, 4 or 5 toppings selected from leven possible choices. That number is
11!/6!5! + 11!/7!4! + 11!/8!3! + 11!/9!2! + 11 + 1. The "1" represents no toppings at all.
That number is 462 + 330 + 165 + 55 + 11 = 1 = 1024. When two pizzas are made, the number of possible combinations for two (including duplicates) is (1024)62 = 1,048,576

Any differences from that number are probably the result of excluding duplicate pizzas.

That (1024)62 should have been (1024)^2. I didn't hit the shift key hard enough

thank you i got it earlier i wasn't seeing a step...

Ah, math problems... the perfect excuse to bring out the clown in me! Let's see if we can make this more fun.

The child calculated 1,048,576 possibilities for the toppings on the two pizzas. Well, that's a lot of pizzas! But according to the answer in the back of the book, they missed a couple. Only 1,048,574 combinations? They were so close! Maybe they counted one pizza with no toppings and one pizza with all toppings – that's a real time-saving move right there!

Now, let's move on to the letter from Joseph F. Heiser. He argued that the two combinations described in part a should be counted as the same. Well, we can't ignore Joseph's opinion, can we? So, if the order of the two pizzas doesn't matter, there are a total of 524,800 possibilities. You see, it's a magic trick! We just cut the number in half – abracadabra!

And just a little note from our dear author: That (1024)62 should have been (1024)^2. Oops, clumsy fingers! It happens to the best of us. So, remember, always hit that shift key with full force when you're in mathematical action!

I'm glad you got it earlier, even with that little step missing. Math can be a sneaky little friend sometimes. But hey, you did it! Enjoy your imaginary pizzas with all the possible toppings. Bon appé!

To solve the problem, you need to calculate the number of possible combinations of toppings for the two pizzas. Let's break down the steps to find the answer to each part of the problem:

a. To verify the child's calculation, you need to calculate the number of possibilities for the toppings on the two pizzas. Little Caesars has 11 toppings to choose from, and you need to consider the order of the two pizzas.

Using the formula for combinations, you can calculate the number of possibilities for each pizza separately and then multiply them together to find the total number of combinations.

For each pizza, you can choose 0, 1, 2, 3, 4, or 5 toppings from the 11 available. You can calculate the number of combinations for each case using the binomial coefficient formula: nCr = n! / (r!(n-r)!).

So, for each pizza:
- 0 toppings: 11! / (0!(11-0)!) = 1 combination
- 1 topping: 11! / (1!(11-1)!) = 11 combinations
- 2 toppings: 11! / (2!(11-2)!) = 55 combinations
- 3 toppings: 11! / (3!(11-3)!) = 165 combinations
- 4 toppings: 11! / (4!(11-4)!) = 330 combinations
- 5 toppings: 11! / (5!(11-5)!) = 462 combinations

The total number of combinations for one pizza is 1 + 11 + 55 + 165 + 330 + 462 = 1024.

Since you have two pizzas, you need to multiply the number of combinations for each pizza together: 1024 * 1024 = 1,048,576.

Therefore, the child's calculation of 1,048,576 possibilities is correct.

b. In this case, you need to calculate the number of possibilities when the order of the two pizzas doesn't matter.

To calculate this, you have to consider that duplicate combinations are allowed because the order doesn't matter.

To find the number of possibilities, you can use the combination with repetition formula: (n + r - 1)Cr = (n + r - 1)! / (r!(n-1)!).

Using this formula, with n = 1024 (the number of combinations for one pizza) and r = 2 (the number of pizzas), you can calculate:

(1024 + 2 - 1)C2 = (1025 - 1)C2 = 1024C2

1024C2 = 1024! / (2!(1024-2)!) = 1024! / (2!1022!) = 1024 * 1023 / 2 = 524,800.

Therefore, when the order of the two pizzas doesn't matter, there are 524,800 possibilities.

It seems like you were missing a step when calculating the number of possibilities. Make sure to include all the combinations for each case and multiply them correctly to get the final result.