Factorial
The factorial of a
number is the product of all the positive integers from 1 upto the number. The
factorial of a given integer n is usually written as n! and n! denotes the product of the first n
natural number.
n! = n x (n – 1) x (n – 2) x …… x 1
n! = n (n – 1)
0! = 1 as a rule.
factorial of a given integer n is usually written as n! and n! denotes the product of the first n
natural number.
n! = n x (n – 1) x (n – 2) x …… x 1
n! = n (n – 1)
0! = 1 as a rule.
Note: Factorial is not defined for improper fractions or negative integers.
npr = n! / (n-r)!
Permutation (factorial !)
If r objects are to be chosen from n, where n is equal or greater then 1, (n ≥ 1) &
these r
objects are to be arranged, and the order of arrangement is
important, then such an arrangement is called a permutation of n objects taken r, at a time.
Permutations is denoted by nPr or (n, r)
nPr = n! x r!
e.g., If it is required to seat 5 men and 4 women in a row such that women occupy the even
places, in how many ways can this be done?
Permutations is denoted by nPr or (n, r)
nPr = n! x r!
e.g., If it is required to seat 5 men and 4 women in a row such that women occupy the even
places, in how many ways can this be done?
m+n ways
5+4 =9 places or positions
In a row of 9 positions, there are four places, and exactly 4 women to occupy them, which is
possible in 4! ways.
In a row of 9 positions, there are four places, and exactly 4 women to occupy them, which is
possible in 4! ways.
The
remaining S places can be filled up by 5 men in 5! ways,
Total number of seating arrangements = 4! X 5! = 24 x 120 = 2880
Important Permutation Rules: n ≥ 1
(i) The total number of arrangements of n things taken r at a time in which a particular thing
always occurs.
e.g., The number of ways in which 3 paintings can be arranged in an exhibition from a set of
five, such that one is always included.
Total number of seating arrangements = 4! X 5! = 24 x 120 = 2880
Important Permutation Rules: n ≥ 1
(i) The total number of arrangements of n things taken r at a time in which a particular thing
always occurs.
e.g., The number of ways in which 3 paintings can be arranged in an exhibition from a set of
five, such that one is always included.
n
number of ways 3. 5-1P3-1 = 3.4P2 = 36
or 3! (4C2) = 6.6 = 36
(ii) The total number of permutations of n distinct things taken r at a time in which a particular
thing never occurs = n-1Pr
e.g., The number of ways in which 3 paintings from a set of five, can be displayed for a photo-
shoot, such that one painting is never picked.
= 5-1P3 = 4P3 ways = 24
It can be observed that
rn-1Pr-1 + n-1Pr = nPr
(iii) The number of permutations of n different objects taken r at a time, when repetitions are
allowed, is nr. The f place can be filled by any one of the n objects in ‘n’ ways. Since repetition is
allowed the second place can be filled in ‘n’ ways again. Thus, there are n x n x n r times ways =
nr ways to fill first r positions. ways = n! x r!
Circular Permutations : Suppose four numbers 1, 2, 3, 4 are to be arranged in the form of a circle.
The arrangement is read in anticlockwise direction, starting from any point as 1432, 4321, 3214
or 2143. (make circle)
These four usual permutation correspond to one circular permutation.
number of ways 3. 5-1P3-1 = 3.4P2 = 36
or 3! (4C2) = 6.6 = 36
(ii) The total number of permutations of n distinct things taken r at a time in which a particular
thing never occurs = n-1Pr
e.g., The number of ways in which 3 paintings from a set of five, can be displayed for a photo-
shoot, such that one painting is never picked.
= 5-1P3 = 4P3 ways = 24
It can be observed that
rn-1Pr-1 + n-1Pr = nPr
(iii) The number of permutations of n different objects taken r at a time, when repetitions are
allowed, is nr. The f place can be filled by any one of the n objects in ‘n’ ways. Since repetition is
allowed the second place can be filled in ‘n’ ways again. Thus, there are n x n x n r times ways =
nr ways to fill first r positions. ways = n! x r!
Circular Permutations : Suppose four numbers 1, 2, 3, 4 are to be arranged in the form of a circle.
The arrangement is read in anticlockwise direction, starting from any point as 1432, 4321, 3214
or 2143. (make circle)
These four usual permutation correspond to one circular permutation.
Thus circular permutations are different only when the relative order of objects to be arranged is
changed.
Each circular permutation of n objects corresponds to n Linear permutations depending on
where (of the n positions) we start.
This can also be though of as keeping the position of one out of n objects fixed and arranging
remaining n – 1 in (n – 1)! ways.
Permutation n ≥ 1 ( r object to be chosen from n & order of r objects of arrangement is important)
Arrangement
Combinations n ≥ r or r ≤ n ( r object to be chosen from n & order of r object is not important)
Selection of comities, teams, pairs…..
If r objects are to be chosen from n, where n is equal or greater then r, r ≤ n and the order of selecting the r objects is not important then such a selection is called a combination of n objects taken r at a time and
denoted by
In a permutation the ordering of objects is important while in a combination it is immaterial. e.g.,
AB and BA are 2 different Permutations but are the same combination.
Usually (except in trivial cases) the number of permutations exceeds the number of
combinations. Trivial cases are when r = 0 or 1.
e.g., If there are 10 persons in a party, and if every two of them shake hands with each other,
how many handshakes happen in the party? n= 10, r = 2 10 > 2 or 2 <10
SoIn: When two persons shake hands it is counted as 1 handshake and not two hence here we
have to consider only combinations.
2 people can be selected from 10 in 10C2 ways.
Hence, number of handshake = 10C2 n!/ (n-r)!r!
1 2 3 4 5 6 7 8 9 10
Combinatorial Identities:
1. nCr = nCn – r
2. nCo = nCn = 1
3. n+1Cr = nCr + nCr – 1
4. n+1Cr+1 = nCr+1 + n–1Cr + n–1Cr-1
5. nPr = r! nC
6. The total no. of combinations of ‘n’ things taken some or all at a time nc = nC1 + nC2 + ……
nCn = 2n – 1
Important Combination Rules
1. The number of combinations of ‘n’ things taken ‘r’ at a time in which p particular thin will
always occur = n-pCr-p P things are definitely selected in 1 way. The remaining r – p things can
be selected from n – p things in n-pCr-p ways.
In how many ways can 7 letters be selected from the alphabet such that the vowels are always
selected.
Soln : There are 5 vowels a, e, i, o, u which are selected in 1 way then possible number of ways
= 26-5C7-5 = 21C2
The number of combinations of ‘n’ things taken ‘r’ at a time in which ‘p’ particular things never
occur is n-pCr (n – p ≥ r)
p things are never to be selected.
Hence r things are to be selected from n - p in n–pCr ways It is clear that n - p ≥ r for this to be
possible.
e.g. In how many ways can 7 letters be selected from the alphabet such that the vowels are
never selected. Vowels = 5 Total consonants = 26
Soln : As vowels (a, e, i, o, u) are never selected. The 7 letters can be selected from (20 – 5),
letters in = 26–5C7 = 21C7 = 116280
3. The number of ways of dividing (partitioning) •n distinct things into r distinct groups, such that
some groups can remain empty = rn
One object ran be put into r partitions in r ways
n objects can be partitioned r x r x r .... n times = rn ways
Examples
i) In how many ways can 11 identical white balls and 9 black bells be arranged in a row so
that no two black balk are together? 11 > or = to 1 and 11> 9
Solution
The 11 white balls can be arranged in 1 way (all are identical)
The 9 black balls can be arranged in the 12 places in 12P9 ways
ii) In how many ways can they be arranged if black balls were identical? (all other conditions
remaining same)
Solution i
The 11 white balls can be arranged in 1 way.
The 9 black balls can be arranged in the 12 places in 12C9 ways.
Thus number of arrangements = 12C9
iii) In how many ways can they be arranged if all the balls are different. (all other conditions
remaining same)
Solution ii
The 11 white balls can be arranged in 11! ways
The 9 black balls can be arranged in the 12 places in 12! ways.
Total number of arrangements = 11!12! / 3!
Example
In a multiple choice test there are 50 questions each having 4 options, which are equally likely.
In how many ways can a student attempt the questions in the test?
Solution
Each question can be attempted in 4 ways and not attempted in 1 way. Each question can be
attempted or unattempted in 5 ways.
Thus 50 questions can be attempted or attempted in 550 ways.
This will include the case when no questions are attempted.
\The student can attempt the paper in (5 to the power of 50) – 1 ways.
Example
How many 4 digit numbers can be formed from the, digits 1, 5, 2, 4, 2, 9, 0, 4, 2
i) with repetition of digits.
ii) without repetition of digits.
Solution
i) In the given set 4 is repeated twice and 2 thrice
\Number of distinct digits = 6
The 4 digit number can be formed in 5.63 ways when repetition is allowed.
Position I can be filled in 5 ways, (as it cannot have O)
The remaining 3 positions can be filled in 6 ways each.
Hence number of numbers = 5.63 = 1080
ii). Position I can be filled in 5 ways.
Position II can be filled in 5 ways (it can contain any of 5 digits except the one in position 1
Thus number of such numbers = 5 x 5 x 4 x 3 = 300
A. Prob (A or B) = Prob (A) + Prob (B)
B. Prob (A and B) = Prob (A) * Prob (B)
C. Prob (A) = Prob (B)
D. Prob (A) + Prob (B) = 1
Correct answer : A.
In mutually exclusive events, the probability of at least one of them occurring
will be equal to the sum of their individual probabilities.
2.
After Wayne Gretzky announced his retirement from hockey, he still had some
regular season games to play. Each member in his family wanted to see him play
one last time. Given that he has 28 family members, and could only get 7
tickets per game, how many different groups of 7 are possible? 28 > 7
a.
Is this a Permutation or a Combination problem?
(Circle one).
Answer : Permutation n ≥
1 nPr = n! / (n – r)!
b. Solve the above problem, Show your work.
Answer
:It will in the form of nPr = 28 P 7
nPr = n! / (n-r)!
28P7 = 28 ! /(28-7)!
= 28! / 21!
= (28 x 27 x 26 x 25 x 24 x 23 x 22 x 21!) /
(21! )
= 28 x 27 x 26 x 25 x 24 x 23 x 22
=
5967561600
3.
If 22 people arrive at a convention together, how many hand-shakes are needed
in order for everyone to meet each other?
(Hint: It takes 2 people to shake hands)
Combination
Answer
nCr =n!/[(n-r)!r!} = 22C2 = 22! / [
(22-2)! 2!] = 231
4. An
upcoming street party (palmerfest) had the following beverages in a large metal
tub: 12 Coca-Cola, 12 Mountain Dew, 8 Evian bottled water, 24 Labatt’s Beer,
and 4 Seagram’s wine coolers. Answer the following questions. Show all
of your work! Leave all answers in decimal form.
Answers
Total Bottles= 12+12+8+24+4 =60 bottles in Tub
a.
If Allan wants a bottled water, what is the probability that the first beverage
he randomly grabs is a bottled water?
Answer : Bottled water / Total Bottles = 8/60 = 0.133
b.
Angie wants a Seagram’s wine cooler. If the beverage taken out of the tub by
Allan was in fact bottled water, what is the probability that the beverage
Angie randomly grabs will be a Seagram’s wine cooler?
Answer: Since Allan got bottled water , there will be in tub only 60-1=59
bottles,
Now if Angies grabs Seagram wine
cooler,
The probability will be
=(Total Seagram’s wine cooler
bottles / Total bottles remaining)
= 4/59 = 0.068
c. Barry.is a thirsty beer drinker. What is the probability that Barry will
select three
Labatt’s
beer consecutively without replacement?
(Start
with the original 60 beverages)
Answer : Probability (24/60)(23/59)(22/58) = 12144/205320 = 0.059
5.
The Ford Model T sold for $950 in 1915 (when the CPI was 10.1). How.much would the ModelT cost in 1995 when adjusted for inflation?
TheCPIin 1995 was 152.4. Show your work.
Answer 1995 price =
1915 price x( 1995 CPI / 1915CPI)
= $950 x ( 152.4 / 10.1)
= 14334.653 $
6.
Below are the frequencies of men and women who enjoy NHL playoff hockey on
television. Use this information to answer the next few questions. Show your
work.
Enjoy NHL Hockey Do Not Enjoy NHL Hockey TOTAL
Men 52 27 79
Women 30 41 71
TOTAL 82 68 150
Joint probability
table
Enjoy
NHL Hockey
|
Do
Not Enjoy NHL Hockey
|
Marginal
Probabilities
|
|
Men
|
52/150 =
0.347
|
27/150 =
0.18
|
0.347+0.18=0.527
|
Women
|
30/150 =
0.20
|
41/150 =
0.273
|
0.20+0.273=0.473
|
Marginal
probabilities
|
0.347+0.20
= 0.547
|
0.18+0.273
= 0.453
|
1
|
a.
What is the probability of selecting at random an individual who is male?
. Answer: From the table
the probability of selecting an individual who is male is 79/150 = 0.527
b. What is the probability of selecting at
random an individual who enjoys NHL hockey?
Answer: From the table the probability of selecting at random an individual who
enjoys NHL hockey is
82/150 =
0.547
c.
What is the probability of selecting a woman given that she does not enjoy NHL
hockey?
Answer:
The probability of selecting a woman given
that she does not enjoy
NHL hockey is
P( Women | Does not enjoy NHL hockey)
=P(Women and Does not enjoy NHL
hockey) /P(Does not enjoy NHL hockey)
=0.273/0.453 = 0.603
d.
What is the probability that a person selected at random enjoys NHL hockey
given that they are male?
Answer:
The probability that a person selected at
random enjoys NHL hockey given that they are male is
P(enjoys NHL hockey | they are male)
=P(enjoys NHL hockey and male)/P( male)
= 0.347 / 0.527 = 0.658
No comments:
Post a Comment