Principles of Mathematics 12: Explained! 284

[Pages:10]Principles of Mathematics 12: Explained!

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Permutations and Combinations

Lesson 2, Part One: Basic Combinations

Basic combinations: In the previous lesson, when using the fundamental

counting principal or permutations, the order of items to be arranged mattered. If all you want to do is select items, and don't care what order they're in, you can use combinations.

Example 1: A committee of 4 people is to be formed from a group of 9 people. How many possible committees can be formed?

This q ue stio n is a c o mb ina tio n sinc e o rd e r is no t imp o rta nt.

The a nswe r is: 9C 4 = 126.

Example 2: A pizza can have 3 toppings out of a possible 7 toppings. How many different pizza's can be made?

The re a re 7 to p p ing s in to ta l, a nd b y se le c ting 3, we will ma ke d iffe re nt typ e s o f p izza .

The way to do questions like this is to use the nCr

feature on your calculator.

n is the total number of items. r is the number of items you want to choose.

For Example 1, you would type into your TI-83: 9 ? Math ? PRB ? nCr ? enter ? 4

This q ue stio n is a c o mb ina tio n sinc e ha ving a d iffe re nt o rd e r o f to p p ing s will no t ma ke a d iffe re nt p izza .

The a nswe r is: 7C 3 = 35.

Questions:

1) How many ways can you select 17 songs for mix CD out of a possible 38 songs?

2) If an ice cream dessert can have 2 toppings, and 9 are available, how many different topping selections can you make?

3) If there are 17 randomly placed dots on a circle, how many lines can you form using any 2 dots?

4) A committee of 4 people is to be formed from a pool of 13 people. How many different committees can be formed?

5) If there are 15 dots on a circle, how many triangles can be formed?

Answers:

1) 38C 17 = 2.878?1010 2) 9C 2 = 36 3) 17C 2=136 4) 13C 4 = 715 5) 15C 3 = 455

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Permutations and Combinations

Lesson 2, Part two: specific items

Combinations including specific items: Sometimes you will be forced

to include or exclude particular items when making a combination. This will reduce the number of items in your selection pool, and also the number of items you can select.

Example 1: A school committee of 5 is to be formed from 12 students. How many committees can be formed if John must be on the committee?

If Jo hn must b e o n the c o mmitte e , yo u' ll ha ve 11 stud e nts re ma ining , o ut o f whic h yo u c a n c ho o se 4.

= 330

Example 2: From a deck of 52 cards, a 5 card hand is dealt. How many distinct five card hands are there if the queen of spades & the four of diamonds must be in the hand?

If the q ue e n o f sp a d e s a nd the fo ur o f d ia mo nd s must b e in the ha nd , yo u' ll ha ve 50 c a rd s re ma ining o ut o f whic h yo u a re c ho o sing 3.

= 19600

Questions:

1) There are 45 songs, and you want to make a mix CD of 18 songs that must include 3 particular songs. How many different selections could you make?

2) If a committee of 7 people is to be formed from a pool of 12 people, but Rachel & Megan must be on the committee, how many selections can be made?

3) There are 9 possible toppings for a sandwich, but you only want 4 toppings, one of which must be pickles. How many different sandwiches can be made?

4) A lottery has 47 numbers, and you must pick 7. How many different combinations are possible if your lucky number 8 must be on each ticket?

5) There are 8 parents and 43 students going on a school trip. Two groups are made, a large one with 30 students and 5 parents, and a small group with 13 students and 3 parents.

a) How many different ways can the parents be chosen for the small group?

b) How many ways can the students be chosen for the large group if Stefan and Dylan must be in the small group?

c) How many ways can students be chosen for the small group if Wade & both his parents must be in the small group?

Answers:

1) 42C 15 = 9.867?1010 2) 10C 5 = 252 3) 8C 3= 56 4) 46C 6 = 9366819 5) a ) 8C 3 = 56

b ) 41C 30 = 3.16?109 c ) 42C 12 = 1.106?1010

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Permutations and Combinations

Lesson 2, Part three: selection pools

Combinations From Multiple Selection Pools: When presented with

multiple groups of items from which you are required to make a selection, you will MULTIPLY the separate cases together.

Example 1: A committee of 3 boys and 5 girls is to be formed from a group of 10 boys and 11 girls. How many committees are possible?

O ut o f the 10 b o ys, we must c ho o se 3: 10C3 O ut o f the 11 g irls, we must c ho o se 5: 11C5

=55440

Example 2: From a deck of 52 cards, a 7 card hand is dealt. How many distinct hands are there if the hand must contain 2 spades and 3 diamonds?

The re a re 13 sp a d e s, we must inc lud e 2: 13C2

13C2 ? 13C3 ? 26C2 =7250100

The re a re 13 d ia mo nd s, we must inc lud e 3: 13C3

Sinc e we c a n' t ha ve mo re tha n 2 sp a d e s a nd 3 d ia mo nd s, the re ma ining two c a rd s

must b e p ulle d o ut fro m the 26 re ma ining c lub s & he a rts. 26C2

Questions:

1) There are 5 meats and 9 veggies available to make a sandwich. How many sandwiches have 2 meats and 6 veggies?

2) How many committees can be formed from 11 men & 9 women if 3 men and 3 women must be on the committee?

3) How many 7 card hands are possible if all the kings must be in the hand?

4) How many 13 card Bridge hands are possible if there are 3 queens, 2 tens, and 2 aces?

5) If there are 19 rock songs and 20 pop songs, how many different ways can you select 12 rock and 8 pop songs for a mix cd?

6) If a crate of radio controlled cars contains 10 working cars and 4 defective cars, how many ways can you take out 5 cars and have only three work?

7) If a student must select two courses from Group A, two courses from Group B, and one course from group C, how many combinations are there?

Answers:

1) 5C 2 ? 9C 6 = 840 2) 11C 3 ? 9C 3 = 13860 3) 4C 4 ? 48C 3 = 17296

4) 4C 3 ? 4C 2 ? 4C 2 ? 40C 6 =552726720

Use 40C 6 sinc e w e c a n' t ha ve a ny q ue e ns, te ns, o r a c e s in the re m a ining six c a rd s.

5) 19C 12 ? 20C 8 = 6347376360 6) 10C 3 ? 4C 2 = 720 7) 4C 2 ? 2C 2 ? 3C 1= 18

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Permutations and Combinations

Lesson 2, Part four: at least/at most

At Least / At Most: These questions will require you to ADD all the

possible cases together. Know the shortcuts too!

Example 1: A committee of 5 people is to be formed from a group of 4 men & 7 women. How many possible committees can be formed if at least 3 women are on the committee?

If a t le a st thre e wo me n a re o n the c o mmitte e , tha t me a ns we c a n ha ve a c o mmitte e with 3 wo me n, 4 wo me n, o r 5 wo me n. Find the c o mb ina tio ns fo r e a c h c a se se p a ra te ly, the n a d d the m a ll to g e the r.

3 Women:

Seven women to choose from, and we require 3: 7C3

Four men to choose from and we require 2: 4C2

Total Combinations: 7C3 ? 4C2

4 Women:

Seven women to choose from and we require 4 7C4

Four men to choose from and we require 1 4C1

Total Combinations: 7C4 ? 4C1

5 Women:

Seven women to choose from and we require 5 7C5

Four men to choose from and we require 0 4C0 = 1

Total Combinations: 7C5

= 371

Example 2: From a deck of 52 cards, a 5 card hand is dealt. How many distinct hands can be formed if there are at least 2 queens?

We c o uld a p p ro a c h this q ue stio n the sa me wa y a s the la st o ne , b ut le t' s use a sho rtc ut inste a d .

At Least/At Most = Total Cases ? Unwanted Cases

The sho rtc ut wo rks sinc e we ha ve the sa me numb e r o f c a rd s in e a c h ha nd , a nd the unre stric te d c o mb ina tio ns fo r a 5 c a rd ha nd must inc lud e e ve ry p o ssib le c o mb ina tio n yo u c a n g e t! Simp ly sub tra c t tho se yo u d o n' t wa nt, a nd yo u' ll b e le ft with the o ne s yo u d o wa nt.

The to ta l p o ssib le c a se s wo uld b e a 5 c a rd ha nd with no re stric tio ns: 52C5

The unwa nte d c a se s a re : no que e ns (O ut o f 48 no n-q ue e n c a rd s, we g e t 5) 48C5 only 1 que e n (O ut o f 4 q ue e ns we g e t 1, a nd o ut o f 48 no n-q ue e ns we g e t 4) 4C1?48C4

= 108336

Example 3: From a deck of 52 cards, a 7 card hand is dealt. How many distinct hands can be formed if there are at most 6 black cards?

The to ta l p o ssib le c a se s wo uld b e a 7 c a rd ha nd with no re stric tio ns: 52C7

The unwa nte d c a se is: 7 blac k c ards (The re a re 26 b la c k c a rd s, a nd we g e t 7) 26C7

= 133126760

No tic e ho w the sho rtc ut ta ke s wa y le ss time tha n a d d ing up a ll the c a se s yo u d o wa nt.

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