SOLUTION: Please show how you got the answer. (a) UMUC Stat Club is sending a delegate of 2 members to attend the 2018 Joint Statistical Meeting in Vancouver, Canada. There are 10 qualified

Question 1111196: Please show how you got the answer.
(a) UMUC Stat Club is sending a delegate of 2 members to attend the 2018 Joint Statistical Meeting in Vancouver, Canada. There are 10 qualified candidates. How many different ways can the delegate be selected?
(b) A bike courier needs to make deliveries at 6 different locations. How many different routes can he take?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
formula for (a) is c(n,x) = n! / (x! * (n-x)!).

n = 10
x = 2

formula becomes c(10,2) = 10! / (2! * 8!).

this can be shown as c(10,2) = (10 * 9 * 8!) / (2! * 8!)

the 8! in the numerator and the denominator cancel out and you are left with:

c(10,2) = (10 * 9) / (2!),

since 2! = (2 * 1), the formula becomes c(10,2) = (10 * 9) / (2 * 1).

simplify to get c(10,2) = 45.

you are using the combination formula of c(n,x) because order is not important.

if order was important, you would be using the permutation formula of p(n,x) = n! / (n-x)!

the difference between the formulas is that c(n,x) is being divided by x! while p(n,x) is not.

the division by x! in the c(n,x) formula removes any considerations of ordering within the set of 2.

an example will help clarify.

suppose you have 3 candidates and you want to select 2 out of the 3.

combination formula says c(3,2) = 3! / (2! * 1!) = 3.

permutation formula says p(3,2) = 3! / 1! = 6.

assume your candidates were a, b, and c.

the 3 sets based on the combination formula would be:

ab, ac, bc

the 6 sets based on the permutation formula would be:

ab, ac, bc, ba, ca, cb

notice that the 6 sets from the permutation formula contain the same members as the 3 sets from the combination formula.

the difference is that the combination formula assumes ab and ba are members of the same set, only in a different order and, since order is not important, combines them into the same set, while the permutation formula says that order is important within the same set and therefore keeps them as separate sets.

for example, you want 2 people to be part of the same set with no other restrictions.

a is one of them and b is the other one.

it doesn't matter who is the leader and who is the follower.
they are both members of the same set and therefore one set is counted.

now you want 2 people, except one of them is the leader and the other is the follower.

now, it's important who is the leader and who is the follower, and you have two sets.

the first set is with a as the leader and the second second set is with b as the leader.

in the first case, you didn't care who was the leader and who was the follower.

in the second case, you did care who was the leader and who was the follower.

that's the difference.

first case order didn't matter.
second case order did.

problem (a) didn't care about the position or order of the members in the delegate team, therefore the combination formula was used.

in problem (b), the problem states:

(b) A bike courier needs to make deliveries at 6 different locations. How many different routes can he take?

he's got 6 locations.

while there might be an optimal route that he can take, the possible number of route can be a very large number.

here you have the same number of locations, but order is important because you can take different routes to get to each location.

the permutation formula for a single set is simply n!.

n = 6, therefore 6! = 6 * 5 * 4 * 3 * 2 * 1 = the total number of possible routes he can take.

that would be 720 possible routes.

it's hard to show the individual routes with 6 locations, but you can show it a lot easier with 3 locations.

the possible routes with 3 locations would be 3! = 3*2*1 = 6 possible routes.

let the routes be a, b, and c.

the possible routes are therefore:

abc
acb
bac
bca
cab
cba

abc means he goes to location a first, then b, then c.
acb means he goes to location a first, then c, then b.
etc.

notice that order is important here because it's the same 3 locations only in a different order.

once you understand what formula to apply, then you can apply it to the larger problem.

6 locations required 6! which is equal to 6 * 5 * 4 * 3 * 2 * 1.

that's a total of 720 possible routes.

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