quant group 15 gregmat

Choice method

Consider number of options in each “slot”, then multiple by each “slot”.

e.g. deck of cards would be 52×51×50, etc

Permutations vs Combinations 1

Order matters in Permutations

Order doesn’t matter in Combinations

Permutations vs Combinations key words

Permutation: lists, arrangement, ranking, routes, “first, second” etc, position, “in a row/line”

Combinations: group, set, unordered, teams, pairs

Permutations Notation

n!/(n-r)!

N is total number, r is what you choose

Combination notation

n!/r!(n-r)!

Permutations with restrictions

Sometimes a seat or slot is restricted to a person. So there is a limited choice.

Sitting together

If 2 or more people or things must be together, consider them as a block. Use choice method like normal, then multiply by the number of ways that block can be organized.

2 people in a block = multiply final by 2

3 people in a block = 6 ways = multiply final by 6

Subtracting Out

To exclude cases, it can be easier to find the total arrangements and then subtract the cases we don’t want.

E.g. A-F people go to movie, with A and B not able to sit next to each other. 6! Is the total arrangements possible, 2x5! Is the undesired (2 people in 5 chairs) is subtracted to get final

Permutations with repeats

For stuff like words where there are repeats/copies, these slots will be the same. Divide the total letters by each group of repeats.

Ex. Mississippi

I = 4

s = 4

p = 2

11!/(4!4!2!)

Permutations in a circle

(N-1)! For circles because of repeates

Combinatorics pattern

Maximum value of (n r) is where r= n/2. If n/2 is not a integer, then r can be n/2 rounded up or down.

Multiple groups

Find combinations per group, multiply all together

Converting to Letters/Words

In some problems, it helps to convert it to a problem of letters or words.

For example, if we want to find the number of ways of going from (0, 0) to (3, 4) where we can move either north or east one unit at a time, and we want to find the number of distinct routes with minimum length, notice that every valid route would be of the form

NNNNEEE

In other words, 4 Ns and 3 Es. We can then simply solve the problem of finding the number of rearrangements with 4 Ns and 3 Es.

Yes or no

Some can be solved by thinking as yes or no for the options, so 2 choices