Permutations

Tag:

BY THE SYMBOL n! ("n factorial") we mean the product of consecutive numbers 1 through n.


n! = 1 x 2 x 3 x . . . . x (n - 2) x (n - 1) x n

=n x (n - 1) x (n - 2) x . . . .x 3 x 2 x 1

The order of the factors does not matter, whether backwards or forwards.


0! is defined as 1. 0! = 1

(The usefulness of this definition will become clear as we continue.)
Example. 6! = 1 x 2 x 3 x 4 x 5 x 6 = 720 Any factorial less than n! is a factor of n!.

Counting Principles

Counting principles describe the total number of possibilities or choices for certain selections. The two fundamental counting principles are listed below.

Fundamental Counting Principle 1:
If the number of events be n, and the number of outcomes for each event in an experiment be Ti (such that i = 1 for the first event, 2 for the second event, ..., and n for the nth event), then the total number of outcomes for all event is T1 x T2 x . . . . x Tn.
Example:
A movie theater sells 3 sizes of popcorn (small, medium, and large) with 3 choices of toppings (no butter, butter, extra butter). How many possible ways can a bag of popcorn be purchased?
Answer:
3 x 3 = 9

Fundamental Counting Principle 2:
If the number of mutually exclusive (no common elements) experiments be m, and the total number of outcomes for all events in each experiment be xj (such that j = 1 for the first experiment, 2 for the second experiment, ..., and m for the mth experiment), then the total number of outcomes for all experiments is x1 + x2 + ... + xm.



Arrangements Sets (n Permutations m)

Permutation is the selection of subsets from a set of elements when the order of the selected elements is a factor.

BY THE PERMUTATIONS of the letters abc we mean all of their possible arrangements: abc acb bac bca cab cba

There are 6 permutations of three different things

The arrangements of n elements taken m at a time represent, in fact, a partial permutation. Older books use the notation nPm (n Permutation m) for arrangements:

nPm = n! / (n-m)!

The number of permutations of n different things taken n at a time is n!.


Important Notes:

a) Number of permutations of n things, taken r at a time, when a particular thing is to be always included in each arrangement = rn-1Pr-1

(b) Number of permutations of n things, taken r at a time, when a particular thing is fixed: = n-1 Pr-1

(c) Number of permutations of n things, taken r at a time, when a particular thing is never taken: = n-1 Pr.

(d) Number of permutations of n things, taken r at a time, when m specified things always come together = m! x ( n-m+1) !

(e) Number of permutations of n things, taken all at a time, when m specified things always come together = n ! - [ m! x (n-m+1)! ]

(f) The number of permutations of n elements with n1 of repeated element, n2 of another repeated element, etc. taken n at a time.
P=n!/(n1!)(n2!)

(g) The number of arrangement of n different things, taken r at a time, when each may be repeated any number of times in each arrangement, is nr



Circular permutations

There are two cases of circular-permutations: -


(a) If clockwise and anti clock-wise orders are different, then total number of circular-permutations is given by (n-1)!

(b) If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by (n-1)!/2!

Number of circular-permutations of n different things taken r at a time:-

(a) If clock-wise and anti-clockwise orders are taken as different, then total number of circular-permutations = nPr /r

(b) If clock-wise and anti-clockwise orders are taken as not different, then total number of circular permutation = nPr/2r

Examples

Q1. Your state issues license plates consisting of letters and numbers. There are 26 letters and the letters may be repeated. There are 10 digits and the digits may be repeated. How many possible license plates can be issued with two letters followed by three numbers?
a. 25000
b. 67600
c. 676000
d. 250000
Ans:
26 x 26 x 10 x 10 x 10

Q2. How many different signals can be made by 5 flags from 8-flags of different colours?
Ans. Number of ways taking 5 flags out of 8-flage = 8P5
= 8!/(8-5)!
= 8 x 7 x 6 x 5 x 4 = 6720

Q3. Find the total number of words that can be made by using all the letters from the word ‘MACHINE’ using only once.
a. 7320
b. 5040
c. 720
d. 12340
Ans:
5040

Q4. In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?
a. 4! x 3!
b. 3! x 3!
c. (4! x 3!)/3!
d. (4! x 3!)/2
Answer:
d. (4! x 3!)/2

Q5. In how many different ways, can 4 boys and 4 girls can arrange themselves in a circle such that boys & girls alternate?
Answer: After fixing one girl round the table, remaining 3 girls can sit in 3!ways. Since boys & girls have to alternate there will be four places, one place each between two girls. These four places can be filled by 4 boys in 4! ways. Hence, by the law of multiplication, the required Number. of ways = 4! x 3! = 144

Q6. How many words can be made by using the letters of the word "SIMPLETON" taken all at a time?
Ans. There are ‘9’ different letters of the word “SIMPLETON”
Number of Permutations taking all the letters at a time = 9P9
= 9! = 362880.

Q7. How many different 4-dgit numbers can be made from the first 4 natural numbers, using each digit only once?
a. 24
b. 18
c. 12
d. 120

Q8. Heather has finally narrowed her clothing choices for the big party down to 3 skirts, 2 tops and 4 pair of shoes. How many different outfits could she form from these choices?
a. 12
b. 24
c. 36
d. 48
Ans: 3x2x4

Q9. In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?
a. 5C3
b. 5P3
c. 53
d. 35
Answer: d

Ryan (not verified)
Ryan 's picture
Doubts in answers to questions 4 & 5

I have got certain doubts in the answers to questions 4 & 5.Here were my solutions -

Q4) Since AAU are the vowels in ABACUS & they form a group, we have 4 elements to arrange ( AAU,B,C,S). These can be arranged in 4! ways.

Also the elements in AAU, themselves can be arranged in 3! ways.

So the answer is 4! x 3! ways. How does the 2 in the denominator come in picture, since i believe there aren't any common terms.

Q5)The girls & boys can be arranged as follows

    G
    B
    G
    B
    G
    B
    G
    B

.

Thus, the girls can be arranged in 4 places in 4! ways & so can the boys. But,since over here we have considered the starting place to be that of a girl, which can be occupied by a boy, we should multiply the resultant by 2.

The ans thus, is 4! x 4! x 2.

Can anyone help me figure out where i am going wrong(assuming the solutions provided are correct).

Incognito (not verified)
Ryan 's picture
About the Q4, with AAU you

About the Q4, with AAU you have 3!/2! permutations, so, in the end you have 4! permutations with the "letters" AAU,B,C,U,S but you have to account for only the ones in AAU that are not repeated (ie, A1,A2,U == A2,A1,U).

4!x(3!/2!) = (4!x3!)/2! = (4!x3!)/2

pragathi's picture
Offline
Last seen: 8 years 44 weeks ago
Joined: 2008-09-13
hi

for the fourt question we divde it by two because there are two similar vowels a

Incognito (not verified)
Ryan 's picture
 ur analysis of ques 4 is

 ur analysis of ques 4 is wrong & given answer is right because the given three vowels can be arranged in 3!/2! not by 3! because two common terms that is A( REPEATED TWO TIMES)

Sponsered Links

All Rights Reserved. Copyright 2006-10 CAT4MBA.com.