@Anonymous: You’re welcome!
excellent explaination. thanks a lot.
maynnnn thax a lot its very cool and easy .
This was really helpful! Thanks a lot for the explanations!
makes sense, yet I got a question wondering if you can help me solve because it seems it is uding both?
Four different mathematics books and six different physiology books are to be arranged on a shelf. How many different arrangements are possible if
the books in each subject must stand together (permutation since order matters) But do i multiply the answers together? IE 432 * 65432 or add 432+65432
second part of questions is if only the mathematics books stand together? then I have no idea! Help!?
Thnx for the help! I desperately needed this!
[…] easy way to remember the difference between permutations and combinations, two terms which are often used interchangeably but incorrectly: Don’t memorize the formulas, […]
Ah, finally i understood something. Arigato!
In response to 688-@R.Magdela-Its pretty much a basic math problem, so, as you know LCM put simply is the multiplication of all the prime numbers(in pairs or without) that make up the numbers in the problem whose LCM is to be taken, and you might know that x! means 12345*…x , using the same property here we have;
4! = 1234
5! = 12345
6! = 123456
Hence here, as you might notice after sorting the numbers to be used in the LCM, its 6! ie 123456 = 720//xD
Thanks Kalid for your help but I am still having trouble understanding the concept.
I am still trying to understand the formula
There are 3 beats and a rest equaling 4 equal spaces of time all together.
|BBBR| BBRB|BRBB|RBBB|
I want to know How many different ways I can play 3 beats and a rest which equals 4 all together .
According to the definition this is a permutation because order matters.
The formula is C(4,1) = 4!/(3! * 1!) = 4
Now if the formula works I should be able to plug different rhythms and beat patterns into the formula and get all the possible patterns with no repetitions.
For example lets take 5 beats and 1 rest for a total of 6 events in time.
|BBBBBR|BBBBRB|BBBRBB|BBRBBB|BRBBBB|RBBBBB|
There are 6 possible patterns of 5 beats and a rest and now I will use the formula to see if it works.
The formula is C(6,1) = 6!/5!*1! = 720/120 = 6
Next example
Lets take 4 beats and rest for a total of 5 events in time.
|BBBBR|BBBRB|BBRBB|BRBBB|RBBBB|
The formula is C(5,1) = 5!/4!*1! = 120/24 = 5
Next lets take 2beats and 2 rest for a total of 4 events in time
|BBRR|BRRB|RRBB|RBBR|
There are 4 possible patterns of 2 beats and 2 rest
How do I put this into the formula ?
The formula is C(4,2) = 4!/2!*1! = 24/2 = 12
What am I doing or thinking wrong.
@Carlie: Thanks!
Hello Jeffreey, I am not sure if you did managed to understand your question, but if not, I hope this will help you!
First off, lets just focus on the 12 eggs and forget about the bad egg. The question we want to solve first is how many different sets of three eggs are there? Inorder to solve this we use C(12,3). Thus, C(12, 3) = 220. Therefore there are a total of 220 sets of eggs if we were to choose 3 eggs at random.
Now we need to calculate within those 220 sets, how many of the sets contain the bad egg. Inorder to solve this we set the following up. Label each of the eggs. First egg is E1, second egg is E2, and so on, and you would have E1, E2, E3, E4, …, E12. Now we choose the first egg to be the bad one. Hence we would have the following sets: (E1, E2, E3), (E1, E2, E4), (E1, E2, E5) and so on. Notice that there are 11 options for the second egg, and 10 options for the third egg for each set. Since order does not count we can now calculate C(11,2). Thus, C(11,2) = 55. Therefore, there are 55 sets of three eggs that have the bad egg.
Prob=55/220 = 0.25 or 25%
I hope this helped you!
[…] Easy Permutations and Combinations […]
Thanks a lot…i could not pass in mid 1 because of confusion in determining the permutations and combinations
I WAS FEELING SO STUPID WHEN I BEFORE I GOT ON THIS WEBSITE NOW I DONT ANYMORE IM HAPPY I NOW KNOW HOW TO FINALLY DO PERMUTATIONS AND COMBINATIONS AND NOT LOOK LIKE A FOOL IN FRONT OF MY CLASS(AP). YOU SAVED MY GRADE IN MY ALGEBRA CLASS. SO, I WANT TO THANK YOU SO MUCH.
kalid??
Hi Kalid, i tried explaining permutation and combination to my sis, and i could succeed only in confusing her. i’ll try ur approach today. she has exams next week . Hope this works
Thanks !
i hav understood to some dit.neway thankzzz
I have a question. There’s a pool for soccer going around. 9 games every week. 3 options for each game. Win, lose or tie. How many different combinations are there in order to hit all 9 games right on 1 sheet and how would you even go about mapping that out?
It was really helpful and a nice way to differentiate between Permutation and Combination