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SECTION 6.4 COMBINATIONS

Ie: 6 swimmers were in a race, how many different outcomes arepossible if 3 swimmers are chosen for 1st, 2nd , & 3rd place?

Since the order is important, this scenario would be a permutation

However, if 3 swimmers are chosen to form an advanced team,then how many different teams are possible?

In this scenario, among the six swimmers, 3 will be chosen and 3 will not

Another way to approach the question is: how many ways can 3 Y’s and 3 N’s be arranged?

YYYNNN

YYNYNN

YYNNYN

YYNNNY

YNYYNN

YNYNYN

YNYNNY

YNNYYN

YNNYNY

YNNNYY

NNNYYY

NNYNYY

NNYYNY

NNYYYN

NYNNYY

NYNYNY

NYNYYN

NYYNNY

NYYNYN

NYYYNN

WHAT IS A COMBINATION?

Combinations refers to the numbers of ways a group of objectscan be arranged, such that the order is NOT important

ie: Suppose 5 {A, B, C, D, E} people were in a game, how manydifferent ways can 3 people be chosen for 1st , 2nd , and 3rd place?

In this situation, the order is important, so we would use permutations

Ie: Suppose 3 people would be chosen to advance into the next round,how many groups of 3 can be created?

If we list out the permutations, some of them will be the samecombination because they have the same letters

ABC

ACB

BAC

BCA

CAB

CBA

ABD

ADB

BAD

BDA

DAB

DBA

ABE

ACE

ADE

BCD

BCE

BDE

ACD

CDE

AEB

BAE

BEA

EAB

EBA

ADC

CAD

CDA

DAC

DCA

AEC

CAE

CEA

EAC

ECA

AED

DAE

DEA

EAD

EDA

BDC

CBD

CDB

DBC

DCB

BEC

CBE

CEB

EBC

ECB

BED

DBE

DEB

EBD

EDB

CED

DCE

DEC

ECD

EDC

Same combination!!

With combinations, the order is not important, so there are only 10 combinations!

FORMULA FOR COMBINATIONS:

When there are “n” number of objects, with “r” number ofobjects to be chosen, the number of combinations will be:

“r” is the number of objects chosen &

“n – r” is the number of objects NOT chosen

The combination formula is also a permutation divided by “r!”

EX: EVALUATE THE FOLLOWING EXPRESSIONS:

Two ways to do this question

PRACTICE: SOLVE FOR ‘N’

2 methods to go from here: crossmultiply or use factors of 56

Since 56 = 8 * 7, assume n = 8

Both sides are equal, therefore

Note: There are 2 answers

“r” represents the number of spaces

Therefore, r = 3 and 4

EX: A GROUP OF 10 STUDENTS HAS 4 BOYS AND 6 GIRLS. IF 4PEOPLE ARE CHOSEN TO FORM A TEAM, HOW MANY DIFFERENTTEAMS CAN BE CREATED IF THERE MUST BE 2 BOYS.

In this scenario, we arechoosing with two differentgroups: Boys vs Girls

Use the “Bucket method” toseparate the two groups

PRACTICE: IF THERE ARE 7 BOYS & 10 GIRLS AND 6 PEOPLE ARECHOSEN TO FORM A TEAM, HOW MANY DIFFERENT TEAMS CANBE CREATED IF THERE MUST BE 2 GIRLS.

EX: IF THERE ARE 4 BOYS & 6 GIRLS, AND 4 PEOPLE ARE CHOSEN TOMAKE A TEAM, HOW MANY DIFFERENT COMBINATIONS CAN BE CREATED IFTHERE MUST BE ATLEAST 1 BOY IN THE TEAM?

Must consider all the different cases with:1 boy & 3 girls, 2 boys & 2 girls, 3 boys & 1 girl , 4 boys & 0 girls

PRACTICE: A TEAM IS TO BE FORMED WITH 6 TEACHERS. IF THERE ARE 5MATH, 8 SCIENCE, & 3 ENGLISH TEACHERS, HOW MANY DIFFERENTTEAMS CAN BE FORMED IF THERE MUST BE ATLEAST 3 MATH TEACHERS?

Must consider all the different cases with:3 Math & 3 Others + 4 Math & 2 Others + 5 Math & 1 Other