The labels from six different cans of soup have come off. If you were to replace them at random, in how many ways could this be done so that
a) none of the cans will be labelled correctly?
b) at least one of the cans will be labelled correctly?
c) all of the cans will be labelled correctly?
a) None of the cans will be labeled correctly? Well, that's soup-er unlucky! Let's see, if none of the cans are labeled correctly, that means each can has only one option out of 5 remaining labels. Therefore, the number of ways this can be done is 5^6 = 15,625 ways. That's a lot of souper mix-ups!
b) At least one of the cans will be labeled correctly? Ah, a glimmer of hope in this soup-er chaotic situation! To find the number of ways, we can calculate the total number of ways to label the cans (6!), and subtract the number of ways where all the cans are labeled incorrectly (1). So, the number of ways where at least one can is labeled correctly is 6! - 1 = 719 ways. Phew, thank goodness for some soup-er luck!
c) All of the cans will be labeled correctly? Well, what a souper way to end our labeling adventure! Since there's only one correct way to label the cans, there's only one way for all the cans to be labeled correctly. Easy peasy, soup-er squeezy!
To solve this problem, we can use the concept of permutations. A permutation is an arrangement of objects in a specific order. In this case, we want to find the number of ways to arrange the cans so that certain conditions are met.
a) For none of the cans to be labeled correctly, we need to find the number of derangements. A derangement is a permutation where none of the objects are in their original position.
The number of derangements of a set of n objects can be calculated using the formula:
D(n) = n!(1/0! - 1/1! + 1/2! - 1/3! + ... + (-1)^n/n!)
For our case, we have 6 cans, so we can calculate:
D(6) = 6!(1/0! - 1/1! + 1/2! - 1/3! + 1/4! - 1/5! + 1/6!)
Simplifying this expression, we get:
D(6) = 6(1 - 1 + 1/2 - 1/6 + 1/24 - 1/120 + 1/720)
= 6(0.63212)
≈ 3.79
Therefore, there are approximately 3.79 ways to arrange the cans so that none of them will be labeled correctly.
b) To find the number of ways at least one can will be labeled correctly, we can subtract the number of derangements from the total number of permutations.
The total number of permutations can be calculated as:
P(6) = 6!
Subtracting the number of derangements, we get:
P(6) - D(6) = 6! - 3.79
= 720 - 3.79
≈ 716.21
Therefore, there are approximately 716.21 ways to arrange the cans so that at least one of them will be labeled correctly.
c) To find the number of ways all of the cans will be labeled correctly, we can use the formula for permutations without restrictions.
The number of permutations of 6 objects can be calculated as:
P(6) = 6!
Therefore, there are exactly 6! = 720 ways to arrange the cans so that all of them will be labeled correctly.
In summary:
a) Approximately 3.79 ways to arrange the cans so that none will be labeled correctly.
b) Approximately 716.21 ways to arrange the cans so that at least one will be labeled correctly.
c) Exactly 720 ways to arrange the cans so that all will be labeled correctly.