btan219 wrote:

eh. im still really confused. the

veritas confused me even more. i really dont understand what they mean by arranging and not arranging.

1)in the first example, they asked: there are 5 friends but only 3 seats in a row. In how many ways can you make 3 of the 5 friends sit in the 3 seats?

5 combinations can sit in seat 1, 4 combinations can sit in seat 2, and 3 combinations can sit in seat 3.

5x4x3=60

what is the equation being used? i dont see any way a!/b! or a!/(b!(a-b)!) will get 60

2) second example says 5 friends and you have to invite any 3 of them to go with you on a vacation. In how many ways can you do that? they give the answer as 10, they didnt state it but im guess they used 5!/(3!(5-3)!)

but isn't it the same thing as the one above? choosing the first friend, i have 5 to choose from. choosing the second friend i have 4 to choose from. choosing the 3 friend i have 3 to choose from. shouldnt it still be 60?

3) and also, i thought the first problem was a permutation (order matters), so why isn't 5!/3!(5-3)!) used? and isn't the second problem a combination problem, why isn't 5!/3! used?

i know these 2 examples are easy and countable, but i want to be able to know which equations to use, im sure the problems aren't going to be this simple in the gmat. but even though its so simple i really just dont get it at all.

this is really confusing and i feel like if i dont get a basic concept as this, i should just give up. how many problems can you possible get about combinations/permutations on the actual gmat anyways?

You actually get only 2-3 questions on P&C so don't worry too much.

That said, the basics of P&C that you are struggling with are not that hard to master. Take another example:

There are 5 students:A, B, C, D and E

Case 1: You have to select 3 students to make a quiz team.

Note here that all 3 students have the same position - member of the quiz team. If you select A, B and C, they make the same team as that when you select B, C and A. There is no "first" team member, "second" team member etc. All 3 members have the same position. All you have to do here is SELECT a group of 3 students out of 5. This is done in 5!/3!*2! = 10 ways

Case 2: You have to select 3 students and give them gold, silver and bronze medals.

Now, you SELECT 3 students as before but you also have to ARRANGE them in first second and third positions (gold, silver and bronze). The three students will occupy three different positions. You do have "first" position, "second" position etc here. So selection and arrangement is done in 5!/2! = 60 ways.

This is similar to the case where out of 5 friends, 3 are selected and made to sit in a row because each seat is different. You have a "first" seat, "second" seat and "third" seat so A, B, C is different from B, C, A. Here again, you have to select and arrange so you use 5!/2! = 60 (Mind you, the two formulas are a!/(a-b)! and a!/b!*(a-b)!)

For more details, I suggest you to check out our Combinatorics book. It starts with the basics and goes on to discuss a lot of GMAT relevant concepts.

_________________

Karishma

Veritas Prep GMAT Instructor

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