sidrah wrote:

i have struggled with the concepts of permutations and combinations since high school.

i still do not understand when order matters and when it does not. for example,:

How many ways are there to award a gold, silver and bronze medal to 10 contending teams?

does order matter here?

If we name the teams A to J

A B C D E F G H I J

1 2 3 n n n n n n n

within teams A, B and C, does it matter who gets 1 - gold, 2 - silver, 3 - bronze?

Here order matters because the way the medals are distributed varies as the order changes. for ex:

* A (Gold), B (Silver) and C (Bronze), and B (Gold), A (Silver) and C (Bronze) are two different ways. Therefore, order matters in this case.

No of people who can get gold medal = 10

No of people who can get silver medal = 10 - 1 = 9

No of people who can get silver medal = 10 - 1 - 1 = 8

so total number of ways the medals can be given = 10x9x8 = 720

Permutation:

When order matters i.e. called permutation. The formula for the permutation is npr = n!/(n-r)! = 10p3 = 10!/(10-3)! = 10x9x8x7!/7! = 720

sidrah wrote:

what about in:

In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

does order matter here?

is there anyone who is able to explain these concepts to me in a simple manner? ive tried everything to try and figure them out but it seems impossible. and i HAVE to figure out when order matters because the formulae are different where order matters and where it doesnt.

HELP.

This is a case of combination.

When order doesnot matter, that is combination. In combination, we are combining a given number of people. Therefore, it cannot have an order. 3 people in a group is is always a group no matter we arrange them.

For ex: 10 people are divided in a group of 3 each.

no of group (combination) = 10c3 = 10!/(3!)(10-3)! = 30

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