Basics of Counting Techniques

 Counting techniques are the very bases of
being able to find the different probabilities of
events in any kind of situation.

 This is not counting one-to-one but this is
collectively counting all possible ways of a
given instance. (Ex. Counting all 4-digit
numbers whose digits are different among one
another)

 FCP
 Factorial Notation
 Permutation
 Combination

 PROBLEM
 Liza brought 3 different pairs of pants and 4 shirts
in a camp. How many combinations of a shirt and a
pair of pants can she choose from to wear?

 Let‟s name the shirts of Liza as Shirts A, B, C,
and D. Let‟s also name her pants as Pants 1, 2,
and 3.
 Let us now enumerate the possible
combinations that Liza can wear.

 The answer is there are 12 possible
combinations.
 We found the answer by drawing a diagram.
But what if there are large numbers given in
the problem (say, 179 pairs of pants and 83
shirts)?

 If an independent event can occur in m ways,
another independent event can occur in n
ways, and another in p ways, then the total
number of ways that all events can occur
simultaneously is
n(E) = m ∙n∙p ways

 In the given problem,
 n(E) = 4 x 3 = 12 ways

 How many 4 digit numbers are there that have
no single digit repeated in it?

 Let us analyze.
 How many numbers (from 0-9) can fit in the
first digit?

 There are only 9. This is because 0 cannot be
the first digit. If it were, then it would not be a
4-digit number anymore.

 How about the second digit?
 Here, zero can already be included, which
makes 10 possibilities.
 HOWEVER, in the problem, it is stated that
digits must not be repeated.

 Therefore, whatever was used on the first digit
cannot be used on the second digit anymore,
which gives us 9 possibilities.

 How about the third digit?
 Again here, we have ten possibilities, but we
have to put into account what was used in the
previous digits, so that makes 8.

 In the last digit, since 3 digits have been used
already, it is safe to say that there are only 7
possibilities.

 All in all, we have
Digit 1st 2nd 3rd 4th
9 x 9 x 8 x 7
Answer: 4536 numbers

 For every independent event, count carefully
the number of possibilities before multiplying
them.

 How many three letter combinations can you
make with the following conditions
(separate)?
 a. Repetition is allowed
 b. Repetition is not allowed
 c. The first letter must be a vowel with no
repetition of letters
 d. There must be no vowel in the first two letters,
repetition is not allowed and the last letter must be
a vowel.

 The factorial of an integer k is the product of
all integers from 1 to k.
 This is usually denoted as k! read as “k
factorial”

 Examples:
 7! = 7∙6∙5∙4∙3∙2∙1 = 5040
 5! = 5∙4∙3∙2∙1 = 120

 The factorial operation is NOT distributive.
 Ex. (5-3)! ≠ 5! – 3!
 The factorial operation cannot be performed
on non-integer numbers.
 Ex. (2.5)!, (√6)!

 The factorial notation precedes multiplication
and/or division.
 Ex. 5!3! ≠ (5∙3)!
 8! / 2! ≠ (8/2)!
 0! = 1

 n! = n(n-1)!
= n(n-1)(n-2)!
= n(n-1)(n-2)(n-3)…(n-k+1)(n-k)!
 This property is very crucial in making
cancellations in factorial expressions.

 A permutation is an ordered arrangement of
objects.
 It tells us how many possible orders there can
be given a number of objects.
 In permutation, if all objects are distinct, then
they cannot be repeated.

 Example
 How many ways can three books be arranged in a
shelf?
 Possible Arrangements
ABC BCA
ACB CAB
BAC CBA

 Take Note:
 Since order is essential, ABC is different from
ACB, and all other similar instances.

 Permutation is usually denoted by
nPr
read as “the permutation of n objects taken r at
a time”

 The general formula for permutation is
nPr = n(n-1)(n-2)…(n-r+1)
or

 n is the total number of objects
 r is the number of objects in consideration
 In the problem given, r=n
 If r=n, the equation becomes

 How many ways can you choose your top 2
senators from a list of 35 candidates?

 Remember Liza and her clothes? Find the
number of ways she can wear her clothes if
the camp lasts for three days and she can only
wear each garment once.

 If we were asked how many permutations we
can get if we rearrange the letters of the word
“cat”, we get 3!
 What if we rearrange the letters of the word
“dad”?

 Permutations of the word “dad”
 dad
 dda
 add

 Actually, there are still 3! permutations for the
word “dad”. However, since we cannot
distinguish the first „d‟ from the second, then
“dad” with the first „d‟ first will be the same as
“dad” with the second „d‟ first.
 This leaves us with 3 distinguishable
permutations.

 The formula for finding the distinguishable
permutations from n set of objects is
where a, b, and c are the number of times a
particular object exists on n.

 Example
 Find the number of distinguishable permutations of
the word “technicalities”

 How many ways can 4 people line up?
 The answer is 4P4 or 4! = 24 ways
 Let us enumerate them.

ABCD BACD CABD DABC
ABDC BADC CADB DACB
ACBD BCAD CBAD DBAC
ACDB BCDA CBDA DBCA
ADBC BDAC CDAB DCAB
ADCB BDCA CDBA DCBA

 Now, how many ways can these 4 people sit
on a round table?

ABCD BACD CABD DABC
ABDC BADC CADB DACB
ACBD BCAD CBAD DBAC
ACDB BCDA CBDA DBCA
ADBC BDAC CDAB DCAB
ADCB BDCA CDBA DCBA
Look closely at the bold italicized letters. If they
are arranged in a circular manner, they all
resemble the same pattern.

 The number of ways n objects can be arranged
in a circular manner is given by

 A combination is an arrangement of objects
with no respect to order.
 Example.
 Find the possible permutations and combinations
of the word “cat”

 For permutations:
CAT CTA
ACT ATC
TAC TCA
There are 6 permutations for the word “cat”.

 For combinations,
 Since order is regardless, then
CAT = CTA = ACT = ATC = TAC = TCA
which gives us only 1 combination.

 What if we take only two letters from the word
“cat”? How many permutations and
combinations are there?

 For permutations,
 According to the formula, there are 3P2 = 6
permutations, which are
AC AT
CA CT
TA TC

 For combinations,
 Again, order is not essential, so AC=CA,
TC=CT, and AT=TA.
 This gives us only 3 combinations.

 The general formula for a combination is
and it is read “the combination of n objects
taken r at a time”

 Find how many possible combinations there
are of 5 cards when randomly selected from a
standard deck of 52 cards.

 Since in getting 5 cards, order is not essential
(i.e getting K♦ 3♠ K♣ 3♦ K♥ is the same as
getting K♥ K♦ 3♠ 3♦ K♣, and all other
orders), then we use combination.

 Find the number of ways of getting just a pair
when randomly getting three cards from a
standard deck.

 Find the number of ways of getting at least a
pair when randomly getting three cards from a
standard deck.

 In a party, there are 53 guests who shook
hands with one another exactly once. How
many handshakes took place?

Basics of Counting Techniques

More Related Content

What's hot

Viewers also liked

Similar to Basics of Counting Techniques

Basics of Counting Techniques