Counting Subsets
Permutations
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How many arrangements of n = 52 cards?
First card can be any of 52
Second card can be any of the remaining 51
…
Once you have chosen 51 there is only one
choice for the last
• So by the product rule, n ∙ (n-1) ∙ (n-2) ∙ … ∙
2.1 = n!
How Many 4-Letter Words
Using Each Letter at Most Once?
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26 choices for first letter
Only 25 for second letter
24 for third letter
23 for fourth letter
So 26∙25∙24∙23
or 26!/22!
Generalized Product Rule
• Let Q be a set of length-k sequences
• if n1 possible 1st elements, n2 possible 2nd
elements (for each first entry), n3 possible
3rd elements (for each 1st & 2nd entry,...)
then,
• |Q| = n1⋅n2⋅⋅⋅nk
How Many Hands with 5 Cards?
• I.e., how many 5-element subsets of a set
with 52 elements?
• We know there are 52! sequences of 52
cards.
• Each sequence uniquely identifies a set of 5
cards: the first 5
• Many-to-one mapping from sequences of 52
cards to sets of 5 cards!
• map sequence a1a2a3a4…a52 to set
{a1,a2,a3,a4,a5}
• How many different sequences map to the
same set?
• Any way of permuting the first 5 elements
maps to the same set (5! ways)
• For each of those, any way of permuting the
last 47 elements maps to the same set
6any4permutation
7 48 any64permutation
7 48
a1a2 a3a4 a5 a6 a52  {a1,a2 ,a3 ,a4 ,a5 }
Counting Subsets
• By the product rule, 5!∙47! different
sequences map to the same set
• Therefore the number of 5-element subsets
of a set of 52 elements is
52!  52   52 
   
5!47!  5   47 
• The number of ways of picking 5 cards to
include is the same as the number of ways
of picking 47 cards to omit!
“n Choose m”
• The number of m-element subsets of a set
of size n is
n 
 n 
n!
 m   m!(n  m!)   n  m 
Counting Doughnut Selections
From 5 kinds of doughnuts
select a dozen.
let A ::= all selections of
12 doughnuts
00 (none) 000000 00 00
chocolate lemon
sugar
glazed plain
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Counting Doughnut Selections
B::= 16-bit words with four 1’s
0011000000100100
00
00 1
1 000000
000000 1 00
00 100
00
chocolate lemon
sugar
glazed plain
Bijection A↔B so |A|=|B|
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10
• # of 16-bit-strings with 4 1’s = # of ways of
choosing 4 from the set of possible
positions {1, …, 16} =
 16 
 4 
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11
How Many Poker Hands (5 cards)
with 2 Jacks?
•
 48 ways
 3 
•
 4 ways
 2 
of picking three non-jacks
of picking two jacks
• So by the product rule the number of
hands with exactly 2 jacks is
 48   4 
48! 4!
 3  g 2   3! 45! 2! 2!
 103776
FINIS