1

Combination Symbols

A supplement to Greenleaf’s QR Text

Compiled by Samuel Marateck ©2009

2

How many 4-card hands consisting of

1 king and 3 queens can be chosen

from a deck?

3

How many 4-card hands consisting of1 king and 3 queens can be chosenfrom a deck?

Since order does not matter and there arefour kings and four queens in the deck,the answer is:

( 4 1) ( 4

3)

4

What is the meaning of ( 4 1)?

It’s the number of ways we can choose one

thing from four, independent of the order.

It is pronounced “four choose one”.

5

Similarly ( 4 3) is the number of ways we can

choose three things from four independent

of the order. It is pronounced “four choose

three”.

6

In ( 4 1) ( 4

3), why do we multiply the

two?

7

For each king there are three queen pairings.

These are the pairings for the king of spades:

k♠ Q♠ Q♣ Q♥

k♠ Q♠ Q♣ Q♦

k♠ Q♣ Q♥ Q♦

k♠ Q♥ Q♦ Q ♠

8

But there are also k♥, k♣ and k♦. So there

are 16 different combinations, four for each

King.

9

What is the probability of choosing

4-card hands consisting of 1 king and 3

queens from a deck?

10

What is the probability of choosing

4-card hands consisting of 1 king and 3

queens from a deck?

( 4 1) ( 4

3) / ( 52 4)

11

We divide by ( 52 4) since this is the number

of ways we can choose four cards at

random from a deck.

12

Let’s evaluate ( 4 1) ( 4

3) / ( 52 4)

13

( 4 1) ( 4

3) / ( 52 4) is:

16/(52*51*50*49/(4*3*2*1))

=0.00006 or .006%

14

Out of how many hands would you expect

to get this hand?

15

Out of how many hands would you expect

to get this hand?

0.00006 is 6x 10-5 , so in 105 hands you

would expect to get 6 such hands or

in one out of 16,666 hands you would get

this hand.

16

How many 5-card hands can you get that

have three aces?

17

How many 5-card hands can you get that

have three aces?

The number of ways we can choose three

aces is ( 4 3) . How many cards are left in

the deck?

18

How many non-aces are in the deck?

There are 48 non-aces left in the deck and

there are two more cards to choose for our

hand.

19

So there are ( 4 3) ( 48

2) ways we can get

three aces:

4*48*47/2 = 4*47*24 = 4512 ways.

20

What is the probability of getting three

aces in a 5-card hand?

21

What is the probability of getting three

aces in a 5-card hand?

( 4 3) ( 48

2) / ( 52 5) =

4512/((52*51*50*49*48)/(5*4*3*2*1)) =

4512/2598960 = .00174

22

What is the probability of winning the

lottery?

23

What is the probability of winning the

lottery?

There are 54 numbers that you can choose

from; the numbers 1 to 54. You must choose

the five correct numbers independent of

their order. The answer is:

24

P(winning) = 1/( 54 5)

( 54 5) = 54*53*52*51*50/120

1/( 54 5) = 3.16 x 10-7

25

If there are 6 pegs distributed in a circle and

a line is drawn from each peg to each other

peg, how many lines are there?

26

For each peg 5 lines are drawn; but there

are 6 pegs. Since, however, each line

connects two pegs, we are overcounting

by 2, so we must divide by 2.

What is the answer?

27

# of lines is 5*6/2 or 15.

28

Another way of looking at this is:

From the first peg, 5 lines are drawn. From

the second peg, 4 lines are drawn since it

is already connected to the first peg. From

the third peg, 3 lines are drawn, since it is

connected to the first two, and so on,

29

For the six pegs, 5+4+3+2+1 or 15 lines are

drawn. For n pegs n-1 + n-2 + n-3 +..+ 1

lines are drawn. We know what the sum

from 1 to m is.

30

The sum is: m(m+1)/2.

Substituting n-1 for m, the sum from 1 to

n-1 is (n-1)(n-1 +1)/2 =?

31

(n-1)(n-1 +1)/2 = n(n-1)/2 which is the

answer we got before.

Can we do this with combination symbols?

32

If there are 6 pegs distributed in a circle and

a line is drawn from each peg to each other

peg, how many lines are there?

33

There are 6 slots:

. . . . . ..

1 2 3 4 5 6

How many ways can we place two item

in these slots?

34

How many ways can we place two item

in these slots?

The answer is ( 6 2).

For n pegs it’s ?

35

For n pegs it’s ( n 2).

36

How many ways can we choose a 5-card

hand so that no two cards have the same

face values?

37

How many ways can we choose a 5-card

hand so that no two cards have the same

face values?

For the first card we have ( 52 1) ways we

can choose the first card. How many

choices do we have for the second card?

38

How many choices do we have for the

second card?

48, since one face value has been

eliminated. So the number of ways we can

choose the second card is:

39

( 48 1).

The third card is?

40

( 44 1).

So the final answer is:

( 52 1) ( 48

1)( 44 1) ( 40

1) ( 36 1).

What is the probability?

41

P(each card has a different face value) =

( 52 1) ( 48

1)( 44 1) ( 40

1) ( 36 1)

( 52 5)

42

In a class of 25, what is the probability that

two or more people have the same

birthdate?

43

In a class of 25, what is the probability that

two or more people have the same

birthdate?

We will first calculate the probability that no

one has the same birthdate.

44

Given the first person, the probability that the second one has a different birth date is364/365. That the first, second and third ones have different birth dates is:1* 364/365*363/365.

For all 25 people?

45

For all 25 people?

P(different birth dates) =

364*363*362*361…341/36524 = 0.47

46

P(2 or more have same birth dates) = .53

47

There are 25 people to be chosen for a

Committee or 5. What is my probability of

being chosen?

48

What is the probability of my being chosen?

( 1 1) ( 24

4)/ ( 25 5).

49

An urn contains 10 red balls and 40 black

ones. What is the probability you will draw

2 red balls.

50

( 10 2) ( 40

0)/ ( 50 2) = 10*9/2 /(50*49/2)

= 45/1225

51

An urn contains 17 red balls and 33 black

ones. What is the probability you will draw

7 red balls if you choose 10 randomly?

52

( 17 7) ( 33

3)/ ( 50 10)

53

A jury pool contains 98 men and 75 women.

12 jurors are chosen at random. What is

the probability that 6 will be women

54

( 98 6) ( 75

6)/ ( 173 12)