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
= 11C4
= 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
= 52C6
= 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 ball
Divide 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 spades
Divide 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 boys
Divide 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 3C2 2C1 2C0 7C3 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
= 3C2 × 2C1 =
×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 = (r1,r2,r3)
6_2R-1W = (r1,r2,w1), (r1,r2,w2), (r1,r3,w1), (r1,r3,w2), (r2,r3,w1), (r2,r3,w2)
3_2R-1B = (r1,r2,b), (r1,r3,b), (r2,r3,b)
3_2R-1G = (r1,r2,g), (r1,r3,g), (r2,r3,g)
3_1R-2W = (r1,w1,w2), (r2,w1,w2), (r3,w1,w2),
1_2W-1B = (w1,w2,b)
1_2R-1G = (w1,w2,g)
6_1R-1W-1B = (r1,w1,b), (r1,w2,b), (r2,w1,b), (r2,w2,b), (r3,w1,b), (r3,w2,b)
6_1R-1W-1G = (r1,w1,g), (r1,w2,g), (r2,w1,g), (r2,w2,g), (r3,w1,g), (r3,w2,g)
3_1R-1B-1G = (r1,b,g), (r2,b,g), (r3,b,g)
2_1W-1B-1G = (w1,b,g), (w2,b,g)
Let :
A : the event of selecting 2 red and 1 white ball
Favorable Choices
6_2R-1W = (r1,r2,w1), (r1,r2,w2), (r1,r3,w1), (r1,r3,w2), (r2,r3,w1), (r2,r3,w2)
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
- 1st sub-event (SE1)
Selecting the items of the first kind
nSE1 : number of ways in which this can be accomplished
- 2nd sub-event (SE2)
Selecting the items of the second kind
nSE2 : 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 ball
Red White Others Total Available 3 2 2 7 To Choose 2 1 0 3 Choices 3C2 2C1 2C0 7C3 - Total Event(E)
Selecting the 2 red and 1 white ball
- 1st sub-event (SE1)
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
⇒ nSE1 = 3C2
- 2nd sub-event (SE2)
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
⇒ nSE2 = 2C1
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
⇒ nE = nSE1 × nSE2
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
⇒ mA = 3C2 × 2C1 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.
- Total Event(E)
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 boys
Men Women Boys Total Available 5 5 10 20 To Choose 2 3 5 10 Choices 5C2 5C3 10C5 20C10 - Total Event(E)
Selecting the 10 members from the available 30
- 1st sub-event (SE1)
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
⇒ nSE1 = 5C2
- 2nd sub-event (SE2)
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
⇒ nSE2 = 5C3
- 3rd sub-event (SE3)
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
⇒ nSE3 = 10C5
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
⇒ nE = nSE1 × nSE2 × nSE3
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
⇒ mD = 5C2 × 5C3 × 10C5 - Total Event(E)
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 | 6C3 | 3C1 | 9C4 |
Alternative 2 | 2 | 2 | 4 |
Choices | 6C2 | 3C2 | 9C4 |
Alternative 3 | 1 | 3 | 4 |
Choices | 6C1 | 3C3 | 9C4 |
- Total Event(E)
Selecting the 5 books from the available 9
- 1st event-alternative (EA1)
- Total Event(E)
Selecting 3 science and 1 math books
- 1st sub-event (SEA11)
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
⇒ nSEA11 = 6C3
- 2nd sub-event (SEA12)
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
⇒ nSEA12 = 3C1
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
⇒ nEA1 = nSEA11 × nSEA12 = 6C3 × 3C1 - Total Event(E)
- 2nd event-alternative (EA2)
Selecting 2 science and 2 math books
Number of ways in which the total event alternative can be accomplished
⇒ nEA2 = nSEA21 × nSEA22 = 6C2 × 3C2 - 3rd event-alternative (EA3)
Selecting 1 science and 3 math books
Number of ways in which the total event alternative can be accomplished
⇒ nEA3 = nSEA31 × nSEA32 = 6C1 × 3C3
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
⇒ nE = nEA1 + nEA2 + nEA3
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)
⇒ mF | = | (6C3 × 3C1) + (6C2 × 3C2) + (6C1 × 3C3) |