# Selecting two or more items from a set of items of varied kinds

# Experiment

## Examples

Selecting two or more

- balls from a container containing three or more balls of different colors
- cards from a pack of playing cards
- cards from a set of three or more numbered cards
- articles from a set of three or more articles of different characteristics
- fruits from a container containing three or more different kinds of fruits

The experiment would be selecting two or more items from the total number of items in consideration.

# Number of possible choices

- Total number of items in consideration
- Number of items being selected

Total number of possible choices

= Number of ways in which the items to be selected can be chosen from the total number of items in consideration

= Number of combinations of total number of items in consideration taking the number of items to be selected at a time

= ^{(Total number of items in consideration)}C_{(Number of items to be selected)}

## Examples

**Experiment**:Drawing 4 balls from a bag containing 5 red and 6 white balls

Total number of balls

= 5 red + 6 white

= 11

Total number of possible choices

= Number of ways in which 4 balls can be drawn from the total 11

= Number of combinations of 11 items taking 4 at a time

=

^{11}C_{4}= 11 × 10 × ... 4 terms 4! = 11 × 10 × 9 × 8 4 × 3 × 2 × 1 = 11 × 10 × 3 = 330 **Experiment**:Drawing 6 cards from a pack of standard playing cards

Total number of cards

= 52

Total number of possible choices

= Number of ways in which 6 cards can be drawn from the total 52

= Number of combinations of 52 items taking 6 at a time

=

^{52}C_{6}= 52 × 51 × ... 6 terms 6! = 52 × 51 × 50 × 49 × 48 × 47 6 × 5 × 4 × 3 × 2 × 1 = 52 × 17 × 10 × 49 × 47

# Number of favorable choices

- Consider the items to be selected for the event to occur favorably
- Divide the total entities into as many groups as the number kinds of items we have to select plus one.
## Examples

**Experiment**:Drawing 3 balls from a bag containing 3 red, 2 white, 1 black and 1 green balls

Let :

**A**: the event of selecting 2 red and 1 white ballDivide the balls into three groups

- Red balls
- White balls
- Others
Consider all the items of the kind other than that we have to select together as others.

Red White Others Total Available 3 2 2 7 **Experiment**:Drawing 3 cards from a standard pack of cards

Let :

**G**: the event of selecting 3 spadesDivide the cards into two groups

- Spades
- Others

Spades Others Total Available **Experiment**:Selecting 10 members for a team

Let :

**D**: the event of selecting 2 men, 3 women and 5 boysDivide the members into four groups

- Men
- Women
- Boys

Men Women Boys Total Available The last group will always be others, consisting of all those kinds which are not to be selected

- Assume that we are selecting as many required from each group
Red White Others Total Available 3 2 2 9 To Choose 2 1 0 3 - The product of the number of ways in which the items from each group can be selected will give us the required number of favorable choices
Red White Others Total Available 3 2 2 7 To Choose 2 1 0 3 Choices ^{3}C_{2}^{2}C_{1}^{2}C_{0}^{7}C_{3}Number of favorable choices in selecting 2 red and 1 white ball

= Number of ways in which 2 red balls can be chosen from the available 3 × Number of ways in which 1 white ball can be chosen from the available 2

= ^{3}C_{2}×^{2}C_{1}=

×3 × 2 2 × 1 2 1 = 3 × 2 = 6 We consider combinations since the order in which the 2 red balls and the white ball are picked is immaterial

First

PickSecond

PickThird

Pick1

2

3red

red

whitered

white

redwhite

red

redAll the above would be interpreted as picking 2 red balls and a white ball

## Check

The above effort is to find the number of favorable choices using mathematical calculations instead of using the set theoretic approach.We can use the set theoretic approach, construct the sample space and derive the same results. However, its use becomes inappropriate as the number of items becomes large.

**Experiment** :

Drawing 3 balls from a bag containing 3 red, 2 white, 1 black and 1 green balls

Sample Space

1_3R = (r_{1},r_{2},r_{3})

6_2R-1W = (r_{1},r_{2},w_{1}), (r_{1},r_{2},w_{2}), (r_{1},r_{3},w_{1}), (r_{1},r_{3},w_{2}), (r_{2},r_{3},w_{1}), (r_{2},r_{3},w_{2})

3_2R-1B = (r_{1},r_{2},b), (r_{1},r_{3},b), (r_{2},r_{3},b)

3_2R-1G = (r_{1},r_{2},g), (r_{1},r_{3},g), (r_{2},r_{3},g)

3_1R-2W = (r_{1},w_{1},w_{2}), (r_{2},w_{1},w_{2}), (r_{3},w_{1},w_{2}),

1_2W-1B = (w_{1},w_{2},b)

1_2R-1G = (w_{1},w_{2},g)

6_1R-1W-1B = (r_{1},w_{1},b), (r_{1},w_{2},b), (r_{2},w_{1},b), (r_{2},w_{2},b), (r_{3},w_{1},b), (r_{3},w_{2},b)

6_1R-1W-1G = (r_{1},w_{1},g), (r_{1},w_{2},g), (r_{2},w_{1},g), (r_{2},w_{2},g), (r_{3},w_{1},g), (r_{3},w_{2},g)

3_1R-1B-1G = (r_{1},b,g), (r_{2},b,g), (r_{3},b,g)

2_1W-1B-1G = (w_{1},b,g), (w_{2},b,g)

Let :

**A** : the event of selecting 2 red and 1 white ball

Favorable Choices

6_2R-1W = (r_{1},r_{2},w_{1}), (r_{1},r_{2},w_{2}), (r_{1},r_{3},w_{1}), (r_{1},r_{3},w_{2}), (r_{2},r_{3},w_{1}), (r_{2},r_{3},w_{2})

# Use of Fundamental Counting Theorem of Multiplication

The total event of selecting the two or more kinds of items forming the favorable choices for the event is sub divided into as many sub events as the number of different kinds of items to be selected

**Total Event(E)**Selecting all the items that make up the favorable choice for the event

**1**^{st}sub-event (SE_{1})Selecting the items of the first kind

n

_{SE1}: number of ways in which this can be accomplished**2**^{nd}sub-event (SE_{2})Selecting the items of the second kind

n

_{SE2}: number of ways in which this can be accomplished- ...
- ...

## Examples

**Experiment**:Drawing 3 balls from a bag containing 3 red, 2 white, 1 black and 1 green balls

Let :

**A**: the event of selecting 2 red and 1 white ballRed White Others Total Available 3 2 2 7 To Choose 2 1 0 3 Choices ^{3}C_{2}^{2}C_{1}^{2}C_{0}^{7}C_{3}**Total Event(E)**Selecting the 2 red and 1 white ball

**1**^{st}sub-event (SE_{1})Selecting 2 red balls from the available 3

Number of ways in which this can be accomplished

= Number of ways in 2 red balls can be selected from the total 3

= Number of combinations of 3 items taking 2 at a time

⇒ n

_{SE1}=^{3}C_{2}**2**^{nd}sub-event (SE_{2})Selecting 1 white ball from the available 2

Number of ways in which this can be accomplished

= Number of ways in 1 white ball can be selected from the available 2

= Number of combinations of 2 items taking 1 at a time

⇒ n

_{SE2}=^{2}C_{1}

Number of ways in which the total event can be accomplished

= Number of ways in which the first sub-event can be accomplished × Number of ways in which the second sub-event can be accomplished

⇒ n

_{E}= n_{SE1}× n_{SE2}Which gives,

Number of ways in which two red balls and a white ball can be picked

= Number of ways in which two red balls can be selected from the available 3 × Number of ways in which a white ball can be selected from the available 2

⇒ m _{A}= ^{3}C_{2}×^{2}C_{1}## Independent sub events

The fundamental counting theorem of multiplication readsIf an event can be subdivided into two or more sub-events which are independent ...

For applying the fundamental counting theorem of multiplication, it is a requirement that the sub events should be independent.

The two sub events of selecting the 2 red balls and selecting the 1 white ball are assumed to be independent. Which white balls can be picked is not influenced by which of the three red balls have been picked and vice versa.

**Experiment**:Selecting 10 members for a team from 5 men, 5 women, 10 boys

Let :

**D**: the event of selecting 2 men, 3 women and 5 boysMen Women Boys Total Available 5 5 10 20 To Choose 2 3 5 10 Choices ^{5}C_{2}^{5}C_{3}^{10}C_{5}^{20}C_{10}**Total Event(E)**Selecting the 10 members from the available 30

**1**^{st}sub-event (SE_{1})Selecting 2 men from the available 5

Number of ways in which this can be accomplished

= Number of ways in which the 2 men can be selected from the available 5

= Number of combinations of 5 items taking 2 at a time

⇒ n

_{SE1}=^{5}C_{2}**2**^{nd}sub-event (SE_{2})Selecting 3 women from the available 5

Number of ways in which this can be accomplished

= Number of ways in which 3 women can be selected from the available 5

= Number of combinations of 5 items taking 3 at a time

⇒ n

_{SE2}=^{5}C_{3}**3**^{rd}sub-event (SE_{3})Selecting 5 boys from the available 10

Number of ways in which this can be accomplished

= Number of ways in which 5 boys women can be selected from the available 10

= Number of combinations of 10 items taking 5 at a time

⇒ n

_{SE3}=^{10}C_{5}

Number of ways in which the total event can be accomplished

= Number of ways in which the first sub-event can be accomplished × Number of ways in which the second sub-event can be accomplished × Number of ways in which the third sub-event can be accomplished

⇒ n

_{E}= n_{SE1}× n_{SE2}× n_{SE3}Which gives,

Number of ways in which 2 men 3 women and 5 boys can be selected

= Number of ways in which 2 men can be selected from the available 5 × Number of ways in which 3 women can be selected from the available 5 × Number of ways in which 5 boys can be selected from the available 10

⇒ m _{D}= ^{5}C_{2}×^{5}C_{3}×^{10}C_{5}

# Use of Fundamental Counting Theorem of Addition

Where an event has alternatives, the fundamental counting theorem of addition is made use of in finding the total number of favorable choices for the event which would be the sum of the favorable choices for each event alternative.

**Experiment** :

Selecting 5 books from 6 science and 3 math books

Let :

**F** : the event of selecting the books such that there would be at least 1 math book

Event F can be accomplished in 3 alternative ways by selecting

- 3 Science and 1 Math books
- 2 Science and 2 Math books
- 1 Science and 3 Math books

Science | Math | Total | |
---|---|---|---|

Available | 6 | 3 | 9 |

To Choose | |||

Alternative 1 | 3 | 1 | 4 |

Choices | ^{6}C_{3} | ^{3}C_{1} | ^{9}C_{4} |

Alternative 2 | 2 | 2 | 4 |

Choices | ^{6}C_{2} | ^{3}C_{2} | ^{9}C_{4} |

Alternative 3 | 1 | 3 | 4 |

Choices | ^{6}C_{1} | ^{3}C_{3} | ^{9}C_{4} |

**Total Event(E)**Selecting the 5 books from the available 9

**1**^{st}event-alternative (EA_{1})**Total Event(E)**Selecting 3 science and 1 math books

**1**^{st}sub-event (SEA_{11})Selecting 3 science books from the available 6

Number of ways in which this can be accomplished

= Number of ways in which the 3 science books can be selected from the available 6

= Number of combinations of 6 items taking 3 at a time

⇒ n

_{SEA11}=^{6}C_{3}**2**^{nd}sub-event (SEA_{12})Selecting 1 math book from the available 3

Number of ways in which this can be accomplished

= Number of ways in which 1 math book can be selected from the available 3

= Number of combinations of 3 items taking 1 at a time

⇒ n

_{SEA12}=^{3}C_{1}

Number of ways in which the total event alternative can be accomplished

= Number of ways in which the first sub-event can be accomplished × Number of ways in which the second sub-event can be accomplished

⇒ n _{EA1}= n _{SEA11}× n_{SEA12}= ^{6}C_{3}×^{3}C_{1}**2**^{nd}event-alternative (EA_{2})Selecting 2 science and 2 math books

Number of ways in which the total event alternative can be accomplished

⇒ n _{EA2}= n _{SEA21}× n_{SEA22}= ^{6}C_{2}×^{3}C_{2}**3**^{rd}event-alternative (EA_{3})Selecting 1 science and 3 math books

Number of ways in which the total event alternative can be accomplished

⇒ n _{EA3}= n _{SEA31}× n_{SEA32}= ^{6}C_{1}×^{3}C_{3}

Number of ways in which the total event can be accomplished

= Number of ways in which the first event alternative can be accomplished + Number of ways in which the second event alternative can be accomplished + Number of ways in which the third event alternative can be accomplished

⇒ n_{E} = n_{EA1} + n_{EA2} + n_{EA3}

Which gives,

Number of ways in which 4 books with at least 1 math book can be selected

= Number of ways in which 3 science and 1 math book can be selected + Number of ways in which 2 science and 2 math book2 can be selected + Number of ways in which 1 science and 3 math books can be selected

= (Number of ways in which 3 science books can be selected from the available 6 × Number of ways in which 1 math book can be selected from the available 3) + (Number of ways in which 2 science books can be selected from the available 6 × Number of ways in which 2 math books can be selected from the available 3) + (Number of ways in which 1 science books can be selected from the available 6 × Number of ways in which 3 math book2 can be selected from the available 3)

⇒ m_{F} | = | (^{6}C_{3} × ^{3}C_{1}) + (^{6}C_{2} × ^{3}C_{2}) + (^{6}C_{1} × ^{3}C_{3}) |