Transcript 3.8 Combinations

3.8 Counting Techniques:
Combinations
• If you are dealt a hand in poker (5 cards), does
it matter in which order the cards are dealt to
you?
A K J 10 Q
A J 10 Q K
10 Q K A J
K J 10 Q A
• Are these hands the same or different?
• The same: order doesn’t matter
Combinations
• Permutation:
– Arrangement of r objects (out of n) in which
the order matters
– E.g.: choosing first, second, third place
• Combination:
– Number of ways of choosing r objects from a
set of n when order doesn’t matter
– E.g.: choosing a group of 3 people
What’s the pattern?
• Choosing 3 objects from, A, B, C, D, E
ABC
ABD
ABE
ACD
BCA
BDA
BEA
CDA
ACB
ADB
AEB
ADC etc.
• There are P(5,3) ways of arranging 3
objects from 5
• There are 3! ways of arranging those three
objects
• However, those 3! ways are “identical”
– CAB is just ABC in a different order
P(5,3)
• So we have
ways to choose 3 objects
3!
from 5:
P(5,3)
5!

3!
(5  3)!3!
 10
There are 10 ways
to choose 3 objects
from a set of 5.
Combinations
#permutations
#combinations =
#permutations of objects selected
• We say “n choose r”
 n
• We write C(n, r) or  
r
 n
n!
 
 r  (n  r )!r !
n
Note: it is NOT  
r
Example 1a
• How many ways can you choose a president
and vice-president from a group of 5
people?
• Does order matter?
• Yes!
• We use permutations
5!
P(5, 2) 
(5  2)!
 20
There are 20 ways to
choose a president and
vice-president from a
group of 5 people.
Example 1b
• How many ways can you choose a
committee of 2 from a group of 5 people?
• Does order matter?
• No!
• We use combinations
 5
5!
There are 10 ways to

 
choose a committee of 2
2
(5

2)!2!
 
 10
from a group of 5
people.
Example 2
You can also use the nCr button on your
calculator.
• How many ways can we choose 30 objects
from a group of 100?
100  100!


 30  70!30!
Type: 100 nCr 30 =
Error on the
calculator!
C (100,30)  2.94 1025
Example 3
How many ways can you choose a hand of 5
cards from a regular deck…
a) with no restrictions?
b) so that exactly one card is an ace?
c) so that all 5 are hearts?
d) so that you have 4 of a kind?
Example 3 sol’ns
a) no restrictions:  52   52!
 
5
47!5!
 2598960
b) exactly one card is an ace:
 4
First we choose the ace:  
1
 48 
Then we choose the rest of the cards:  4 
So the number of ways is  4  48   778320
 1  4 
Example 3 sol’ns
c) all 5 are hearts:
How many hearts? 13 Choose 5 of those cards:
 13 
   1287
5
d) 4 of a kind:
Choose the value first:
 13 
 
1
Then choose the last card:
 4
Then choose the suits:  
 4
 48 
 
1
So number of ways to get 4 of a kind:
 13   4   48 
    
 1  4  1 
 624
0!
• What is 0! ?
• We define 0! = 1
• How many ways can you select 0 people
from a group of 4?
• 1
 4
4!
 
• Check:
 0  4!0!
24

24(1)
1
Types of Reasoning
• When working out problems, two types of
reasoning can be used
• Direct reasoning
– All suitable outcomes are totaled
• Indirect reasoning
– All undesired outcomes are subtracted from
total
• Why would we do this?
• Sometimes the calculations are easier!
Example 4
• In how many ways can you pick 5 people
from a group of 6 adults and 8 children if
the group must contain at least 2 adults?
• This means we have a group with
–
–
–
–
exactly 2 adults
or a group with exactly 3 adults
or a group with exactly 4 adults
or a group with exactly 5 adults !
• Direct:
Example 4 solution
# ways to have at least 2 adults = # ways to have
exactly 2 adults + # ways to have exactly 3 adults +
etc.
 6   8   6   8   6  8   6  8 
                
 2   3   3   2   4  1   5  0 
 1526
• Indirect:
# ways to have at least 2 adults = total # of groups of
5 – # groups with no adults – # of groups with 1
adult
 6  8   6  8 
 14         
    0  5   1  4 
5


 1526
Example 5
• What is the probability of a specific set of
numbers winning Lotto 6/49?
• We choose 6 numbers from a set of 49
n( A)
n( A)  1
P( A) 
 49 
n( S )
n( S ) 

1
 49 
 
6
1

13 983 816
 
6
For those of you
planning your future, this
is very, very small.
Poker
• What is the probability of being dealt the
following hands:
– 1 pair (2 of a kind, 3 different)?
– 2 different pairs (2 of a kind, 2 of a different kind,
1 different)?
– 3 of a kind (3 the same and 2 different)?
– Straight? (a run of consecutive values, at least one
card is a different suit (A can be high or low))
– Flush? (all five cards are of the same suit, but not
all consecutive)
Poker
• What is the probability of being dealt the
following hands:
– Full house? (1 pair and 3 of a kind)
– 4 of a kind?
– Straight flush? (a run of cards of the same suit,
but not 10, J, Q, K, A)
– Royal flush? (10, J, Q, K, A of the same suit)
– None of the above?