PREMUTATIONS/COMBINATIONS - the Difference. (1 Viewer)

[dolbinau]

Yes. I hate extension. and I hate this topic, along with circle geometry and binomial theorem. But that's another thing

Anyway, what is the difference?

How do I know when to use 8C5 or 8P5. Can people give me some tips.

If I was selecting 5 people from a group of 8, is that permutation or combination?

 

[Larry123]

Combination's are used when arrangements of objects are not relevant
whereas Permutations are used when arrangements of objects are relevant.

 

[aakash]

Its combinations.

Basically, permutation is when order IS important and combination is when order IS NOT important (you probably heard this many times, so consider this example)
Also, for question which has words like 'select', 'combination', 'choose','unordered' use combination.
similarly for words like 'arrange', 'permutation', 'ordered'- use permutation.
(this usually works)

imagine I am choosing 3 people out of 6 (a,b,c,d,e,f)
now if i choose a,b,c or a,c,b (ie in different order), its still the same three people so order does not matter and hence its a combination question.

on the other hand if I have to make a 3 digit number out of 6 digits (1,2,3,4,5,6) - and i choose 1,2,3 (in that order), it corresponds to 123
whereas if i choose then in 1,3,2 (in that order), it correspond to 132 which is a different number and should be counted seperately (and hence its permutation).

Hope that helped.

 

[dolbinau]

Yes, it did. Thanks.

 

[bored of sc]

If I was selecting 5 people from a group of 8, is that permutation or combination?

Combinations since you are selecting them, the order in which you select them does not matter/is irrelevant.

 

[blonde brain]

im really bad at permutations and combinations. how do i answer this question (step by step)

There are 3 identical blue marbles and four identical yellow marbles in a row. How many different arrangements are possible? How many different arrangements of just 5 of these marbles are possible?

 

[lolokay]

the first part of the question is equivalent to saying "how many ways can you choose 3 marbles from 7 marbles to be made blue?" which is 7C3 = 7!/3!4!

alternatively, arrange the marbles in a row (7!) then divide out by 3! to unorder the blue marbles, and divide out by 4! to unorder the yellow marbles (unordering because their order doesn't matter, as they're all the identical). you do this type of thing when you're finding the number of arrangements of the letters of a word for example

for the second part, I think you need to go by cases - ie how many cases would there be if there were 1/2/3 blue marbles?
So it becomes 5C1 + 5C2 + 5C3

 

[Da fishaks]

Hey all u need to know when to use "p" or "c" is tht with p order counts although with c order doesnt... So if ur slecting 5 people from a group of 8 it wil be 8c5 as order doesnt count. Either way i hate permutations and combinations anywayz...:S

 

[lolokay]

I mostly think of it as ordering (multiply by n!) and unordering (divide by n!) rather than P and C

+also, perm/comb = best 3u topic

 

[gurmies]

^ Agreed =)

 

[grasshopper1]

well we started permutations today and i cant say its the best thing ive done, a few things i could do with some help

1) how do you approach the questions, as in sometimes i struggle to understand what i have to do to get to the answer?

2) is there any way to remember how to approach them?

example.
11. A queue has 4 boys and 4girls standing in a line. Find how many diffeent arrangments of the line are possible if
a) the boys and girls can stand anywhere
b)the boys and girls alternate
c)2 particular girls wish to stand together
d)all the boys stand together
e)also find the probability that 3 particular people will be in the queue together if the queue forms randomly

thanks

 

[lolokay]

a) 8! -> just order the 8 people
b)4!*4!*2 -> order each of the 4 girls and boys, multiply by 2 since either girls or boys could be in the odd/even spots
c) first view the 2 girls as one person -> this gives 7! ways. but the 2 girls can order themselves in 2! ways, so 7!*2!
d) now, we firstly view all the boys as one person -> this gives 5! ways to order, and then 4! to order the boys amongst themselves, so 4!*5!
e) view the 3 people as one person, gives 6!*3! ways. but there are a total of 8! ways to arrange the people in the line. so probability = 6!3!/8! = 3/28

 

[grasshopper1]

thanks mate yeah ive done the the questions now, i guess the best way to get good at them is just keep doing them

 

[Timothy.Siu]

yeah thats the way to go