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0018 DATA ANALYSIS, PROBABILITY and STATISTICS
A permutation tells us the number of ways we can combine a set where {a, b, c} is
different from {c, b, a} and without repetition.
•
•
•
r is the size of of the group you are picking.
n is the size of the sequence from which you are choosing your objects
! is the a factorial operator where, N!= Nx(N-1)x(N-2)x(N-3)x…x2x1.
Example
Take the first 10 letters of the alphabet and find out how many groupings of 5
letters can be made without repetition.
n= 10
r=5
10!
(10 − 5)!
10 x9 x8 x7 x6 x5 x 4 x3 x 2 x1
= 10 x9 x8 x7 x6 = 30240
5 x 4 x3 x 2 x1
A combination is a set of unique elements where order does not matter. For example
{a, b, c} is the same as {c, b, a}.
•
•
•
k is the size of of the group you are picking.
n is the size of the sequence from which you are choosing your objects
! is the a factorial operator where, N!= Nx(N-1)x(N-2)x(N-3)x…x2x1.
Example
A club has 16 members. The club is choosing a committee of 6. How many
different committees can be chosen?
n=16
k=6
16!
16 x15 x14 x13 x12 x11x10 x9 x8 x7 x6 x5 x 4 x3 x 2 x1
=
6!(10)!
6 x5 x 4 x3 x 2 x1(10 x9 x8 x7 x6 x5 x 4 x3 x 2 x1)
=
16 x15 x14 x13 x12 x11
= 4 x14 x13 x11 = 8008
6 x5 x 4 x3 x 2
A sequence is a pattern of numbers in a definite order following a certain rule. Usually
denoted as a1,a2,a3,a4,… an. A sequence can either finite or infinite. It may have a set
number of terms or it may be endless like the Fibonacci sequence shown below.
Arithmetic Sequences-An arithmetic sequence is a progression in which each term
(except the first term) is obtained from the previous term by adding a constant known as
the common difference.
The general form of an arithmetic sequence:
a, a + d, a + 2d, a + 3d, …
where d equals the common difference.
Formula to find the nth term of an arithmetic sequence:
a + (n-1)d
Example
7, 10, 13, 16, 19, 22, 25,…
a = 7 the common difference d = 10 – 7 = 3
The eighth term in the sequence would be
25 + 3 = 28
or
7 + ( 8 – 1 ) x 3 = 7 + 21 = 28
Geometric Sequences- A geometric sequence is a progression in which you get the next
term by multiplying the previous term by a non-zero constant, which we call the common
ratio.
The general form of a geometric sequence:
a, ar, ar2, ar3, ... where r equals the common ratio.
Formula to find the nth term of a geometric sequence:
arn-1
Example
3, 6, 12, 24, 48, 96
a= 3
the common ratio r = 6 / 3 = 2
The seventh term in this geometric series would be
96 x 2 = 192
or
(7-1)
6
3x2
= 3 x 2 = 3 x 64 = 192
Fibonacci Sequence is a sequence that occurs often in nature, as in the spirals of a
pinecone or the arrangement of sunflower seeds. It named by Leonardo Da Vinci when
he discovered it by watching the breeding habits of rabbits:
a0=0 and a1=1 and an= an-1 + an-2
The nth Fibonacci number can be found by this formula:
n
1
5
n
1+ 5 − 1− 5 ]
) (
)
2
2
[(
Finite Differences are found by subtracting two consecutive numbers in a sequence. The
first order finite differences of a sequence can be found by subtracting all of the
consecutive terms. The second order finite differences of a sequence can be found by
subtracting the consecutive terms of the first order differences.
Example
Given the sequence {0, 1, 4, 9, 16, 25,…}
Sequence
0
1
4
9
16
25
First order finite difference
1
3
5
7
9
Second order finite difference 2
2
2
2
A series is the sum of terms in a sequence. Denoted as a1+a2+a3+ a4+… +an. A series can
also either be finite or infinite depending on the nature of the sequence it is summing.
Arithmetic series- the sum of the terms of an arithmetic sequence.
The sum of the first n numbers of an arithmetic series can be found by.
n/2 [2a + (n - 1) d]
or
n/2 [ first term + last term]
Example
Sequence: 8, 11, 14, 17, 20, 23, 26,…
Series: 8, 19, 33, 50, 70, 93,
The seventh term in the series would be
93 + 26 = 119
or
7/2 [ 2 x 8 + ( 7 – 1 )x 3] = 7/2 [ 16 + 6 x 3] = 7/2 [34] = 119
Geometric Series- the sum of the terms of an arithmetic sequence
The sum of the first n numbers in a geometric series can be found by.
or
Example
Sequence: 3, 6, 12, 24, 48, 96
Series: 3, 9, 21, 45, 93
The sixth term of the series would be
93 + 96 = 189
or
[ 3 ( 26-1) ] / 2 – 1= [ 3 x 63 ] / 1 = 189
Binomial- A polynomial with two unlike terms. (a + b)
Binomial coefficients – The coefficients of the binomial (x + y)n
“Binomial expansions can be completely specified by their coefficients, since the
determination of the exponents is straightforward: In the expansion of (a + b)n, the
exponents of a begin at n and decrease to 0, and the exponents of b begin at 0 and
increase to n. Therefore we can think of the square expansion simply as the coefficients
"1 2 1" and the cube expansion as "1 3 3 1". It turns out that binomial coefficients display
some very interesting relationships. These relationships can be seen effectively by
arranging the coefficients into a triangular array, where the nth row consists of the
coefficients of the expansion of (a + b)n.” (math.Rutgers.edu)
1
1
1
1
2
1
1
3 3
1
4 6 4 1
1 5 10 10 5 1
1
This arrangement is also known as Pascal’s Triangle is a list of the binomial coefficients
in triangular form. Each term can be found by adding the number above and to the left to
the number above and to the right of the term you want.
Let
denote the mth entry of the nth row of Pascal's triangle (where the top row is row
0 and the leftmost "1" of each row is entry 0). It turns out that the number of ways of
choosing m objects from n is exactly
. The reason for this is not too difficult to see:
The number of times that the term an – mbm appears in the expansion of (a + b)n = (a + b)
(a + b) ··· (a + b) is exactly the number of ways to "choose" m powers of b from the n
available factors (a + b).
We will next obtain an explicit formula for the binomial coefficient
. To do this we
will need the factorial function, which multiplies a natural number n by all natural
numbers less than n. It is denoted by n! (read "n factorial"). Thus
(By definition, 0! = 1.) For example, 4! = 4 · 3 · 2 · 1 = 24 and 2! = 2 · 1 = 2. Compute n!
for n = 3, 5, 6, 7, 8, 9, 10.
To derive the explicit formula for
in terms of the factorial function, we work through
the possibilities of choosing m objects from n objects. First let us choose the objects one
at a time. For the first, we have n objects from which to choose. Once we have selected
one, we have n – 1 objects from which to choose a second. After that, we have n – 2
objects from which to choose a third, etc. The total number of ways to choose m objects
in order is therefore
However, we are interested in the number of ways to choose m objects from n objects
without respect to order. Therefore we divide this number by m!, the number of different
orders in which we can select the same m objects. So we have the identity
(math.Rutgers.edu)
from this we get the binomial expansion
Reference List
http://mathworld.wolfram.com
http://mathwords.com/
http://mathforum.org/
http://math.com/
Problems
1) A geometric progression has positive terms. The sum of the first six terms is nine
times the sum of the first three terms. The seventh term is 320. Find
(i) the common ratio
(ii) the first term
(iii) the sum of first five terms
2) The sum of the first 10 terms in an arithmetic progression is 50 and the sum of the next
10 terms is 250. Find the thirteenth term
3) The second term of an arithmetic progression is nine times the fifth term and the sum
of the first eight terms is 56. Find
4) The series log x + log 2 + lgx2 + log 4 + logx3 + log 8 + ... is an arithmetic progression.
Show that the sum of first ten terms is 55 log 2x.
5) Given the 26 and 40 are the 4th and 5th numbers in a sequence and the second finite
difference of the sequence is a constant of 3. Find the first three terms in the sequence.
6) What are the coefficients of the third, fourth, and fifth terms in (x +y)8
7) In how many ways can a sorority of 20 members select a president, vice president and
treasury, assuming that the same person cannot hold more than one office
8) In a regular deck of cards how many different poker hands are there?
Solutions
1) Sum of first 6 terms = 9 (Sum of first 3 terms)
1 - r6 = 9 (1 - r3)
r - 9r3 + 8 = 0
3
(r - 8)(r3 - 1) = 0
r = 2 or r = 1 (rejected as G. P. is increasing)
6
Seventh term,
a.27-1 = 320
a.26 = 320
a=5
Sum to first five terms
= 155
2) Sum of first 10terms = 10/2 [2a + (10 - 1) d] = 50
5 (2a + 9d) = 50
2a + 9d = 10...(1)
Sum of first 20terms = 20/2 [2a + (20 - 1) d] = 250 + 50
10 (2a + 19d) = 300
2a + 19d = 30...(2)
Solving (1) and (2) simultaneously,
d = 2, a = -4
13th term = a +(13 - 1) d
= -4 + 12(2)
= 20
3) Second term = 9 (Fifth term)
a + d = 9 (a + 4d)
8a + 35d =0
.....(1)
Sum to first eight terms:
8/2 [2a + (8 - 1) d] = 56
8a + 28d = 56 .....(2)
Solving (1) and (2) simultaneously,
d = -8, a = 35
Let the least number of terms be n.
n/2 [2 (35) + (-8) (n - 1)] < 0
39n - 4n2 < 0
n (39 - 4n) < 0
n > 9 3/4
since n > 0
The least number of terms is 10.
4) log x + log 2 + logx2 + log 4 + logx3 + log 8 + ...
= (log x + log 2) + 2 (log x + log 2) + 3 (log x + log 2) + ...
power low of logarithms
This is an Arithmetic Progression with
first term = log x + log 2
= log 2x
common difference = log x + log 2
= log 2x
Sum to first 10terms = 10/2 [ 2 log 2x + (10 - 1) log 2x]
= 5 (11 log 2x)
= 55 log 2x
5)
A
B
D
C
26
E
3
F
3
40
G
3
G = 40 – 26 = 14
G – F = 3 F = G – 3 = 14 – 3 = 11
E = F – 3 = 11 – 3 = 8
D=E–3=8–3=5
C = 26 – F = 26 – 11 = 15
B = C – E = 15 – 8 = 7
A=B–D=7–5=2
The sequence is 2, 7, 15, 26, 40
These problems found at:
http://library.thinkquest.org/C0110248/algebra/apgpintro.htm
6)
1
Row
0
1
1
1
1
1
1
1
1
8
3
5
6
7
56
Row
3
4
20
4
Row
1
15
6
21
56
Row
1
5
35
70
2
1
10
35
Row
1
6
15
1
3
10
21
28
2
4
Row
1
5
7
28
Row
1
6
1
8
Row
7
1 8 Row
So the third fourth and fifth coefficients are 28, 56, 70.
Or
C
C
,
C
,
C
28, 56, 70
n k
8 2 8 3 8 4
7)
This problem and more practice can be found at:
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut56_per
m.htm
8) n = 52
k=5
N!
K !( N − K )!
52!
52 x51x50 x 49 x 48
311875200
=
=
= 2598960 possible poker hands
5! x 47!
5 x 4 x3 x 2
120