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.

• 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

1. Consider the items to be selected for the event to occur favorably
2. 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

     1. Red balls
     2. White balls
     3. 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

     1. Spades
     2. 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

     1. Men
     2. Women
     3. Boys

     	Men	Women	Boys	Total
     Available

     The last group will always be others, consisting of all those kinds which are not to be selected

3. 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

4. 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 Pick	Second Pick	Third Pick
   1 2 3	red red white	red white red	white red red

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

Check

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)

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 (SE₁)

  Selecting the items of the first kind

  n_SE1 : number of ways in which this can be accomplished

• 2^nd sub-event (SE₂)

  Selecting the items of the second kind

  n_SE2 : number of ways in which this can be accomplished

• ...
• ...

Examples

1. 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

   • 1^st 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

     ⇒ n_SE1 = ³C₂

   • 2^nd 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

     ⇒ n_SE2 = 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

   ⇒ 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

   ⇒ mA

   =

   3C2 × 2C1

   Independent sub events

   The fundamental counting theorem of multiplication reads

   If 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.

2. 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

   • 1^st 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

     ⇒ n_SE1 = ⁵C₂

   • 2^nd 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

     ⇒ n_SE2 = ⁵C₃

   • 3^rd 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

     ⇒ n_SE3 = 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

   ⇒ 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

   ⇒ mD

   =

   5C2 × 5C3 × 10C5

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

• Total Event(E)

  Selecting the 5 books from the available 9

• 1^st event-alternative (EA₁)
  • Total Event(E)

    Selecting 3 science and 1 math books

  • 1^st 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

    ⇒ n_SEA11 = ⁶C₃

  • 2^nd 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

    ⇒ n_SEA12 = 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

• 2^nd 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

• 3^rd 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

⇒ 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)