Problem Solving With Permutations

In summary: For the problem with distributing the ice cream among 10 children, you would do the following:There are 10 children.There are 4 types of ice cream.Each child can have 1 type of ice cream.Therefore, the total number of ways to distribute the ice cream among the 10 children is 40 ways.
  • #1
koolkidx45
8
0
Hi, I have the following questions

1) A coin is tossed 9 times. In how many ways could the results be six heads and three tails?

2) A man bought two vanilla ice cream cones, three chocolate ones, four strawberry, and one butterscotch. In how many ways could he distribute them among his 10 children?

3) Juan's soccer team played a total of 14 games this season. Their record was eight wins, four losses, and two ties. In how many orders could this have happened?

I think for the above problems you take the total factorial and divide by the duplicate factorials. For ex for 1) 9! / 6! x 3! = 84

Any help is appreciated. Thanks!
 
Physics news on Phys.org
  • #2
To try to explain the formula think about it like this. Let's say the coins were labeled 1 to 9. Let's say we were lining them up on a table and on the table we marked out three slots for tails and six slots for heads.

t t t | h h h h h h

How many ways are there to arrange coins into these slots? Well you have 9 choices for the first slot, 8 for the next, etc. So there are 9! ways.

But now we notice that for example

1 2 3 | 4 5 6 7 8 9

and

2 1 3 | 4 5 6 7 8 9

Correspond to the same coins being heads and the same set of coins being tails.

Okay so say that you have a fixed set of 3 tail-coins, and a fixed set of 6 tail coins. How many permutations does that correspond to? Well line up the tail coins first, there are 3! ways to do that and then there are 6! possibilities for the head coins.

Okay so

TOTAL_PERMUTATIONS = NUMBER_OF_POSSIBLE_SETS * PERMUTATIONS_PER_SET

9! = NUMBER_OF_POSSIBLE_SETS * (6!3!)
NUMBER_OF_POSSIBLE_SETS = (9!)/(6!3!)

So yes your formula is correct.
 
  • #3
Err sorry this was wrong
 
Last edited:
  • #4
welcome to pf!

hi koolkidx45! welcome to pf! :wink:
koolkidx45 said:
I think for the above problems you take the total factorial and divide by the duplicate factorials. For ex for 1) 9! / 6! x 3! = 84

yes, that's correct :smile:

(and for three types, you have three factorials on the bottom instead of two)
 
  • #5
Thanks for all the help guys! So for my two other problems i would do the same thing? But the denominators would be different factorials? For ex.

3) 14!/8!*4!*2!
 
  • #6
yup! :biggrin:
 

FAQ: Problem Solving With Permutations

What is a permutation?

A permutation is an arrangement of objects where the order matters. It is a way of organizing objects or events in a specific sequence.

What is the difference between a permutation and a combination?

A permutation is an ordered arrangement of objects, while a combination is an unordered selection of objects. In other words, the order of objects matters in permutations, but not in combinations.

How do you calculate the number of permutations?

The number of permutations can be calculated using the formula n! / (n-r)! where n represents the total number of objects and r represents the number of objects in each permutation.

What are some real-life examples of permutations?

Some real-life examples of permutations include arranging a deck of cards, setting a password with a specific sequence of numbers, or choosing a committee with a specific order of positions.

How are permutations used in problem solving?

Permutations are used in problem solving to determine the total number of possible outcomes, to calculate the probability of specific events occurring, and to find the most efficient solution to a problem. They are also used in fields such as mathematics, computer science, and statistics to analyze data and make predictions.

Similar threads

Replies
19
Views
2K
Replies
3
Views
1K
Replies
1
Views
4K
Replies
45
Views
4K
Replies
2
Views
3K
Replies
1
Views
3K
Replies
2
Views
7K
Back
Top