Permutations and combinations help!?

Question

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.

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.

Thanks for the help! This is getting complicated.

Answer 1

(a) The combinations are, for number of toppings:
0 -> 11C0 = 11!/11! = 1 combo
1 -> 11C1 = 11!/10!1! = 11 combos
2 -> 11C2 = 11!/9!2! = 55 combos
3 -> 11C3 = 11!/8!3! = 165 combos
4 -> 11C4 = 11!/7!4! = 330 combos
5 -> 11C5 = 11!/6!5! = 462 combos

462 + 330 + 165 + 55 + 11 + 1 = 1,024 combinations from 0 to 5 toppings.

Given 1,024 combinations per pizza:

1,024 (first pizza) * 1,024 (second pizza) = 1,048,576 total combinations

=================
(b) Of the 1,048,576 combinations, there are 1,024 where the second pizza is the same as the first. Those combinations appear only once in the list of 1,048,576.

The remaining 1,047,552 (1,048,576 - 1,024) have differing first and second pizza. Those appear twice (in either order) in the list of 1,047,552.

The number of unique combinations is:

1,024 (pairs not counted twice)
+ (1,047,552 / 2) (half of pairs counted twice)
---------
524,800 unique combinations of two pizzas.

Answer 2

I agree with McFate on (a), but I get a slightly different answer for (b). Mac Fate, I think you have counted some combinations twice.

(b) There are 1024 combinations possible for the first pizza. Given a choice of the first pizza, there are 1023 possibilities for the second pizza which are different from the first. So if the order of number of pizza matters, there are 1024 * 1023 possibilities. If the order does not matter, we have counted each combination twice, so there are 1024 * 1023 / 2 = 523,776 possibilities.

Or, to put it a different way, there are C(1024, 2) = 523,776 possibilities.

Answer 3

permit her % the ribbons one by one. For the 1st one, she has 6 concepts. For the 2d, there are 5 concepts left. So there are 30 distinctive techniques of choosing an ordered pair of ribbons. in spite of the shown fact that, Cathy does not care pertaining to to the order. Taking first a blue and then a white ribbon is comparable to taking a white ribbon first and then a blue one. this suggests we counted double the style of possibilities. So the actually selection is 5*6 / 2 = 15. that's, in effect, a mix: 2C6. The reasoning in the back of the formulation for combinations is as follows: you could positioned all six ribbons in 6! distinctive orderings. Of the six regarded after ribbons you could then opt for the 1st 2. As you do not care pertaining to to the order of those 2, you could divide the selection by the style of techniques of ordering those 2: 2!. an identical is going for the 4 ribbons Cathy did not opt for: those could be in 4! distinctive orderings. So the full type of achieveable combinations is 6! / (2! 4!)