Adv. Algebra

7/24/2019 Adv. Algebra

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Basics Concepts and Formulas in Permutations and Combinations

1. Fundamental Principles of Counting : Multiplication Theorem

If an operation can be performed in m different ways and following which a second

operation can be performed in n different ways, then the two operations insuccession can be performed in m n different ways

2. Fundamental Principles of Counting : Addition Theorem

If an operation can be performed in m different ways and a second independentoperation can be performed in n different ways, either of the two operations can be

performed in (m+n) ways.

3. Factorial

et n be a positi!e integer. "hen n factorial (n#) can be defined as

n# $ n(n%1)(n%2)...1

Examples

i. $ & ' ' 3 ' 2 ' 1 $ 12

ii. 3# $ 3 ' 2 ' 1 $ *

Special Cases

iii. # $ 1

i!. 1# $ 1

. Permutations

ermutations are the different arrangements of a gi!en number of things by taing

some or all at a time

Examples

i. -ll permutations (or arrangements) formed with the letters a, b, c by taing

three at a time are (abc, acb, bac, bca, cab, cba)

ii. -ll permutations (or arrangements) formed with the letters a, b, c by taing

two at a time are (ab, ac, ba, bc, ca, cb)

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&. Combinations

ach of the different groups or selections formed by taing some or all of a number

of ob/ects is called a combination

Examples

i. 0uppose we want to select two out of three girls , , . "hen, possible

combinations are , and . (Note that PQ and QP represent the sameselection)

ii. 0uppose we want to select three out of three girls , , . "hen, onlypossible combination is

*. Difference beteen Permutations and Combinations and !o to Address

a Problem

0ometimes, it will be clearly stated in the problem itself whether permutation orcombination is to be used. owe!er if it is not mentioned in the problem, we ha!e to

find out whether the 4uestion is related to permutation or combination.

5onsider a situation where we need to find out the total number of possible samples

of two ob/ects which can be taen from three ob/ects , , . "o understand if the

4uestion is related to permutation or combination, we need to find out if the order isimportant or not.

If order is important, will be different from , will be different from and

will be different from

If order is not important, will be same as , will be same as and will

be same as

ence,If the order is important, problem will be related to permutations.

If the order is not important, problem will be related to combinations.

6or permutations, the problems can be lie 78hat is the number of permutations thecan be made7, 78hat is the number of arrangements that can be made7, 78hat are

the different number of ways in which something can be arranged7, etc

6or combinations, the problems can be lie 78hat is the number of combinations thecan be made7, 78hat is the number of selections the can be made7, 78hat are the

different number of ways in which something can be selected7, etc.

9ostly problems related to word formation, number formation etc will be related to

permutations. 0imilarly most problems related to selection of persons, formation of

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geometrical figures , distribution of items (there are e'ceptions for this) etc will be

related to combinations.

:. "epetition

"he term repetition is !ery important in permutations and combinations.

5onsider the same situation described abo!e where we need to find out the total

number of possible samples of two ob/ects which can be taen from three ob/ects, , .

If repetition is allowed, the same ob/ect can be taen more than once to mae a

sample.

i.e., if repetition is allowed, , , can also be considered as possible samples.

If repetition is not allowed, then , , cannot be considered as possible

samples

#ormall$ repetition is not alloed unless mentioned specificall$%

;. p4 and 4p are two different permutations ,but they represent the same

combination.

n, nr$

Special Case:n$ 1

nris also denoted by (n,r).nrhas importance outside combinatorics as well where

it is nown as the falling factorial and denoted by (n)ror nr

Examples

i. ;2$ ; ' : $ &*

ii. &$ & ' ' 3 ' 2 $ 12

1. #umber of permutations of n distinct things ta&ing all at a time

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=umber of permutations of n distinct things taing them all at a time $ nn$ n#

11. #umber of Combinations of n distinct things ta&ing r at a time

=umber of combinations of n distinct things taing r at a time ( n5r) can be gi!en by

n

5r$ n!/(r!)(nr)!=n(n1)(n2)(nr+1) /r! where 0rn

If r > n, n5r$

Special Case:n5$ 1

n5ris also denoted by 5(n,r).n5roccurs in many other mathematical conte'ts as well

where it is nown as binomial coefficient and denoted by (nr)

Examples

i. ;52$ 87/21$ 2;

ii. &5$ 5432/4321$ &

'seful "elations in Permutations and Combinations

(% 'seful "elations ) Factorial

n* + n%,n)(-*

2. nCr=nPr /r!

.% 'seful "elations ) Permutations

1. nPn+ n*

2. nP/+ (

3. nP(+ n

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. nPn+nPn ) (

&. nPr+ n,n)(Pr)(-

0% 'seful "elations ) Combinations

1. nCr+nC,n ) r-

'ample

8C6=8C2=8721= 28

2. nCn+ (

3. nC/+ (

. nC/1nC(1

nC21 %%% 1nCn+ 2

n

'ample

i. 5+51+

52+53+

5$ (1 + + * + + 1) $ 1* $ 2

&. nCr)(1nCr+

,n1(-Cr (Pascal's Law)

*. nCr/nCr-1= n-r+1/r

:. 3f nCx+nC$ then either x + $ or ,n)x- + $

&. Selection from identical ob4ects: Some Basic Facts

1. "he number of selections of r ob/ects out of n identical ob/ects is 1

2. "otal number of selections of ?ero or more ob/ects from n identical ob/ectsis n+1.

More Concepts and Formulas: Permutations

1. Permutations of 5b4ects hen All 5b4ects are #ot Distinct

"he number of ways in which n things can be arranged taing them all at a time,

when p1of the things are e'actly alie of 1sttype, p2of them are e'actly alie of a

2ndtype, and prof them are e'actly alie of rthtype and the rest of all are distinct is

n!/p1! p2! ... pr!

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2. Permutations ith "epetition : #umber of permutations of n distinct

things ta&ing r at a time ,"epetition alloed-

"he number of permutations of n distinct things taing r at a time when each thing

may be repeated any number of times is nr

3. Circular Permutations: Case (: hen cloc&ise and anticloc&ise

arrangements are different

=umber of circular permutations (arrangements) of n different things is (n%1)#

. Circular Permutations: Case 2: hen cloc&ise and anticloc&ise

arrangements are not different

=umber of circular permutations (arrangements) of n different things, when

clocwise and anticlocwise arrangements are not different (i.e., when obser!ations

can be made from both sides), is 1/2(n1)!

ermutations under estrictions

1. Case (: 6hen s particular things are ala$s to be included

=umber of permutations of n different things taing r at a time, when s particular

things are always to be included in each arrangement, is(n%s)5(r%s) r#

@eri!ation of the formula

(rAs) ob/ects can be selected from the (nAs) ob/ects in(n%s)

5(r%s)ways.

s ob/ects can be selected from s ob/ects only 1 way.

"otally r ob/ects are selected and these can be arranged in r# ways

"otal number of arrangements $ (n%s)5(r%s) r#

-lternati!e 6orm

0ome te't boos gi!es the formula as

(n%s)

(r%s)

r

swhich is same as

(n%s)

5(r%s) r#

(n%s)5(r%s) r# $ (ns)!r!/[ns(rs)]!(rs)!=(ns)! r!/(nr)! (rs)!

(n%s)(r%s)rs$ (ns)!/[ns(rs)]!] r!/(rs)! = (ns)! r!/(nr)! (rs)!

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-s you can see, both these formulas are actually the same.

2. Case 2: 6hen a particular thing is ala$s to be included

=umber of permutations of n different things taing r at a time, when a particularthing is always to be included in each arrangement, is

(n%1)5(r%1) r#

@eri!ation of the formula is similar to that of case 1.

-lternati!e 6orm-s we ha!e seen for case 1, the same formula can also be e'pressed as (n%1)(r%1) r

3. Case .: 6hen s particular things are ne7er included

=umber of permutations of n different things taing r at a time, when s particular

things are ne!er included is(n%s)5r r#


emo!e the s ob/ects which are ne!er included and we are left with (n%s) ob/ects. rob/ects can be selected from these (nAs) ob/ects in (n%s)5rways.

r ob/ects can be arranged in r# ways

"otal number of arrangements $ (n%s)5r r#

. Case 0: 6hen a particular thing is ne7er included

=umber of permutations of n different things taing r at a time, when a particular

thing is ne!er included, is(n%1)5r r#

@eri!ation of the formula is similar to that of case 3

&. Case 8: 6hen m particular things ala$s come together

=umber of permutations of n different things taing them all at a time, when m

particular things always come together, is(n%m+1)# m#


Broup these m ob/ects and consider it as a single ob/ect. "hen the total numberob/ects $ (n%m+1). "hese (n%m+1) ob/ects can be arranged in (n%m+1)# ways

m ob/ects can be arranged in m# ways

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"otal number of arrangements $ (n%m+1)# m#

*. Case 9: 6hen m particular things ne7er come together

=umber of permutations of n different things taing them all at a time, when m

particular things ne!er come together, is

n# % (n C m + 1)# m#

@eri!ation of the formula"otal number of arrangements possible using n different ob/ects taing all at a time $

n#

=umber of arrangements of n different things taing all at a time, when m particularthings always come together, is

(n%m+1)# m# (As we have seen in case 5)

ence, number of permutations of n different things taing all at a time, when m

particular things ne!er come together $ n# % (n%m+1)# m#

9ore 5oncepts and 6ormulas in 5ombinations

1. Combinations ith "epetition : #umber of combinations of n distinctthings ta&ing r at a time ,"epetition alloed-

=umber of combinations of n distinct things taing r at a time when each thing maybe repeated any number of times is ,n1r)(-Cr

2. Total #umber of Combinations : Case (

"otal number of combinations is the total number of ways of selecting one or more

than one things from n distinct things . i.e., we can select 1 or 2 or 3 or D or n items

at a time.

"otal number of combinations $ n51+n52+ ... +

n5n$ 2n% 1

3. Total #umber of Combinations : Case 2

=umber of ways of selecting one or more than one ob/ects out of 0 1alie ob/ects of

one ind, 02alie ob/ects of the second ind and 03alie ob/ects of the third ind is

(01+ 1) (02+ 1)(03+ 1) % 1

"he general formula for the same can be defined as gi!en below.

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one ind, 02alie ob/ects of the second ind ,03alie ob/ects of the third ind and soon ... 0nalie ob/ects of the n

thind is

(01+ 1) (02+ 1)(03+ 1)...(0n+ 1) % 1

. Total #umber of Combinations : Case .


one ind, 02alie ob/ects of the second ind and rest p different ob/ects is

(01+ 1) (02+ 1)2Ep% 1

"he general formula for the same can be defined as gi!en below.


one ind, 02alie ob/ects of the second ind and so on ... 0 nalie ob/ects of the

nthind and rest p different ob/ects is

(01+ 1) (02+ 1) ... (0n+ 1) 2Ep% 1

5ombinations under estrictions

1. Case (: 6hen s particular things are ala$s to be included

=umber of combinations of n different things taing r at a time, when s particular

things are always to be included in each selection, is(n%s)5(r%s)

2. Case 2: 6hen a particular thing is ala$s to be included

=umber of combinations of n different things taing r at a time, when a particular

thing is always to be included in each selection, is(n%1)5(r%1)

3. Case .: 6hen s particular things are ne7er included

=umber of 5ombinations of n different things taing r at a time, when s particular

things are ne!er included in any selection, is(n%s)

5r

. Case 0: 6hen m particular things ne7er come together

=umber of 5ombinations of n different things taing r at a time, when m particular

things ne!er come together in any selection, isn5r%

(n%m)5(r%m)

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More Formulas in Permutations and Combinations

(% Sum of all numbers formed from gi7en digits

If all the possible n digit numbers using the n distinct digits are formed, the sum of

all the numbers so formed is e4ual to

(n%1)# (0um of the n digits) (111 ... n times)

2% #umber of persons and handsha&es

-ssume that in a party e!ery person shaes hand with e!ery other person.

et n $ the total number of persons present in the party

h $ total number of handshaes

"hen,

h=n(n1) / 2

.% Derangements

-ny change in the e'isting order of things is called a derangement.

If n things are arranged in a row, the number of ways in which they can bederanged so that none of them occupies its original place is

n!(11/1!+1/2!1/3!++(1)n1/n!)=n!r=0n(1)r1/r!

Beometrical 6igures in ermutations and 5ombinations

In this chapter, we will be dealing with the formulas related to permutations and

combinations in!ol!ing geometrical figures.

1. #umber of triangles formed b$ 4oining the angular points of a pol$gon

"he number of triangles that can be formed by /oining the angular points of a polygonof n sides as !ertices are

n(n1)(n2) / 6

2. #umber of uadrilaterals formed b$ 4oining the 7ertices of a pol$gon

"he number of 4uadrilaterals that can be formed by /oining the !ertices of a polygon

of n sides are

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n(n1)(n2)(n3) / 24 where n>3

3. #umber of triangles formed b$ 4oining n points out of hich m points are

collinear

5onsider there be n points in a plane out of which m points are collinear. "he number

of triangles that can be formed by /oining these n points as !ertices are

n53%m53

. #umber of triangles formed b$ 4oining n points out of hich no three

points are collinear

5onsider there be n points in a plane out of which no three points are collinear. "he

number of triangles that can be formed by /oining these n points are

n(n1)(n2) / 6

&. #umber of straight lines formed b$ 4oining n points out of hich m


5onsider there be n points in a plane out of which m points are collinear. "he number

of straight lines that can be formed by /oining these n points are

n52%m52+ 1

*. #umber of straight lines formed b$ 4oining n points out of hich no


5onsider there be n points in a plane out of which no points are collinear. "he number

of straight lines that can be formed by /oining these n points are

n(n1) / 2

:. #umber of rectangles formed b$ using hori;ontal lines and 7ertical lines

"he number of rectangles that can be formed by using m hori?ontal lines and n

!ertical lines are

m52n52

;. #umber of diagonals formed b$ 4oining the 7ertices of a pol$gon

"he number of diagonals that can be formed by /oining the !ertices of a polygon of n

sides are

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n(n3) / 2

Examples:

(%Fut of : consonants and !owels, how many words of 3 consonants and 2 !owels

can be formedG

'planation H

=umber of ways of selecting 3 consonants out of : $ :53=umber of ways of selecting 2 !owels out of $ 52

=umber of ways of selecting 3 consonants out of : and 2 !owels out of $ :53'52

=(765 / 321)(43 / 21)=210

It means that we can ha!e 21 groups where each group contains total & letters(3

consonants

and 2 !owels).

=umber of ways of arranging & letters among themsel!es $

$ & ' ' 3 ' 2 ' 1 $ 12

ence, e4uired number of ways $ 21 ' 12 $ 2&2

2%In a group of * boys and girls, four children are to be selected. In how many

different ways can they be selected such that at least one boy should be thereG'planation H

In a group of * boys and girls, four children are to be selected such that

at least one boy should be there.

ence we ha!e choices as gi!en below

8e can select boys %%%%%%(Fption 1).

=umber of ways to this $ *5

8e can select 3 boys and 1 girl %%%%%%(Fption 2)=umber of ways to this $ *53'

51

8e can select 2 boys and 2 girls %%%%%%(Fption 3)

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=umber of ways to this $ *52'52

8e can select 1 boy and 3 girls %%%%%%(Fption )

=umber of ways to this $ *51'53

"otal number of ways$ (*5) + (

*53'51) + (

*52'52) + (

*51'53)

$ (

*

52) + (

*

53'

51) + (

*

52'

52) + (

*

51'

51) Applied the formula

n

Cr=

n

C(n - r) J

=[65 / 21]+[(654 / 321)4]+[(65 / 21)(43 / 21)]+[64]$ 1& + ; +


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"he word KF"I5-K has : letters. It has the !owels KFK,KIK,K-K in it and these 3 !owels

should always come together. ence these three !owels can be grouped and consideredas a

single letter. "hat is, "5(FI-).

ence we can assume total letters as &. and all these letters are different.=umber of ways to arrange these letters $ $ & ' ' 3 ' 2 ' 1J $ 12

-ll "he 3 !owels (FI-) are different

=umber of ways to arrange these !owels among themsel!es $ 3# $ 3 ' 2 ' 1J $ *

ence, re4uired number of ways $ 12 ' * $ :2

9%In how many ways can a group of & men and 2 women be made out of a total of :

men and 3 womenG

'planation H

8e need to select & men from : men and 2 women from 3 women

=umber of ways to do this$ :5&'

352$ :52'

351Applied the formulanCr=

nC(n - r) J

=(76 / 21)3$ 21 ' 3 $ *3


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Number of ways to arrange these vowels among themselves = 4! / 2!=4321 / 21=12

ence, re4uired number of ways $ 1; ' 12 $ 12


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$ number of ways to arrange the !owels ' number of ways to arrange the consonants

$ * ' * $ 3*

(.%- bag contains 2 white balls, 3 blac balls and red balls. In how many ways can 3

balls be drawn from the bag, if at least one blac ball is to be included in the drawG'planation H

6rom 2 white balls, 3 blac balls and red balls, 3 balls are to be selected such thatat least one blac ball should be there.

ence we ha!e 3 choices as gi!en below

8e can select 3 blac balls %%%%%%%%%%%%%%%%%%%%%%%%%%(Fption 1)

8e can select 2 blac balls and 1 non%blac ball%%%%%%(Fption 2)8e can select 1 blac ball and 2 non%blac balls%%%%%%(Fption 3)

=umber of ways to select 3 blac balls $ 353

=umber of ways to select 2 blac balls and 1 non%blac ball $ 352 ' *51

=umber of ways to select 1 blac ball and 2 non%blac balls $ 351 ' *52

"otal number of ways

$ 353 + (352 ' *51) + (351 ' *52)

$ 1 + (351 ' *51) + (351 ' *52) Applied the formula nCr=nC(n - r) J

=1+[36]+[3(65 / 21)]$ 1 + 1; + &

$ *

(8%In how many ways can the letters of the word K-@K be arrangedG 'planation

H

"he word K-@K has * letters.

Lut in these * letters, KK occurs 2 times and rest of the letters are different.

Hence,number of ways to arrange these letters = 6! /2!=654321 / 21=360

(9%ow many words can be formed by using all letters of the word KLI-KG

'planation H

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"he word KLI-K has & letters and all these & letters are different.

"otal words formed by using all these & letters $ &&$

$ & ' ' 3 ' 2 ' 1 $ 12

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0ince the digits & is placed at unit place and another one digits is placed at tens place,

we ha!e now four digits remaining. -ny of these digits can be placed at hundredsplace.

& 1

e4uired =umber of three digit numbers $ ' & ' 1 $ 2

2/%8hat is the !alue of 12G

'planation H

12$ 1 '


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20. -n e!ent manager has ten patterns of chairs and eight patterns of tables. In how

many ways can he mae a pair of table and chairG

'planation H

e has has 1 patterns of chairs and ; patterns of tables

ence, - chair can be arranged in 1 ways and- table can be arranged in ; ways

ence one chair and one table can be arranged in 1 ' ; ways $ ; ways

29%- bo' contains red, 3 white and 2 blue balls. "hree balls are drawn at random.6ind out the number of ways of selecting the balls of different coloursG

'planation H

1 red ball can be selected in 51ways

1 white ball can be selected in3

51ways1 blue ball can be selected in 251ways

"otal number of ways

$ 51'351'

251$ ' 3 ' 2

$ 2

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'planation H

1stman can draw water from any of the & taps2ndman can draw water from any of the remaining taps

3rdman can draw water from any of the remaining 3 tapsthman can draw water from any of the remaining 2 taps

&thman can draw water from remaining 1 tap

& 3 2 1

ence total number of ways $ & ' ' 3 ' 2 ' 1 $ 12

.(%ow many two digit numbers can be generated using the digits 1,2,3, without repeating

any digitG

'planation H

8e ha!e four digits 1,2,3,

"he first digit can be any digit out of the four gi!en digits

=ow we ha!e already chosen the first digit. 0ince we cannot repeat the digits, we are

left

with 3 digits now. "he second digit can be any of these three digits

3

0ince the first digit can be chosen in ways and second digit can be chosen in 3 ways,

both the digits can be chosen in 3 $ 12 ways. !eference " #ultiplication $heorem%

i.e., 12 two digit numbers can be formed

.2%"here are three places , and such that 3 roads connects and and roads connects and . In how many ways can one tra!el from to G

'planation H

"he number of ways in which one can tra!el from to $ 3 $ 12

!eference " #ultiplication $heorem%

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.0%"here are 1 women and 1& men in an office. In how many ways a team of a man and awoman can be selectedG

'planation H

=umber of ways in which a team of a man and a woman can be selected

$ 1& 1 $ 1& !eference " #ultiplication $heorem%

.8%In how many ways can three boys can be seated on fi!e chairsG

'planation H

"here are three boys.

"he first boy can sit in any of the fi!e chairs (& ways)

&

=ow there are chairs remaining. "he second boy can sit in any of the four chairs (ways)

&

=ow there are 3 chairs remaining. "he third boy can sit in any of the three chairs (3

ways)

& 3

ence, the total number of ways in which 3 boys can be seated in & chairs

$ & 3 $ *

.>%In how many ways can : boys be seated in a circular orderG

'planation H

!eference " Circular Permutations" Case &%=umber of arrangements possible $ (:%1)# $ *# $ * & 3 2 1 $ :2

.=. In how many ways can : beads can be arranged to form a neclaceG

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'planation H

!eference " Circular Permutations" Case " when clocwise and anticlocwisearranements are not different%

=umber of arrangements possible

=1/2(71)!=1/26!=1/2654321=360

0/%In how many ways can a team of & persons can be formed out of a total of 1persons such that two particular persons should be included in each teamG

'planation H

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Solution ( : 'sing the Principles

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

"wo particular persons should be included in each team

i.e., we ha!e to select &%2 $ 3 persons from 1%2 $ ; persons

ence, the re4uired number of ways $ ;53

=876 / 321=87=56

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Solution 2 : 'sing the Formula

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

eference H 5ase 1H 8hen s particular things are always to be includedJ

=umber of combinations of n different things taing r at a time, when s particular things

are always to be included in each selection, is(n%s)5(r%s)

ere n $ 1, r $ &, s $ 2

ence, the number of ways $ (n%s)5(r%s)$;53

=876 / 321=87=56

02%ow many triangles can be formed by /oining the !ertices of an octagonG

'planation H

!eference " Num*er of trianles formed *+ ,oinin the anular points of a pol+on%

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"he number of triangles that can be formed by /oining the angular points of a polygon of

n sides as !ertices are

n(n1)(n2) / 6

ere n $ ;

ence, the number of triangles that can be formed by /oining the !ertices of an octagon

=n(n1)(n2) / 6 = 8(81)(82) /6=8.7.6 / 6=56

0.%If there are < hori?ontal lines and < !ertical lines in a chess board, how many

rectangles can be formed in the chess boardG

'planation H

!eference " Num*er of rectanles formed *+ usin horiontal lines and vertical lines%

"he number of rectangles that can be formed by using m hori?ontal lines and n !ertical

lines are

m52n52

ere m $


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ere n $ 1

ence, "he number of diagonals

=n(n3) / 2=10(103) / 2=107 / 2=57=35

08. 6ind the number of triangles that can be formed using 1 points in a plane such that

points are collinearG

'planation H

!eference " Num*er of trianles formed *+ ,oinin n points out of which m points arecollinear%

5onsider there be n points in a plane out of which m points are collinear. "he number oftriangles that can be formed by /oining these n points as !ertices are

n53%m53

ere n $ 1, m $

ence, "he number of triangles $ n53%m53$

153%53

$ 153%51

n5r$n5(n % r)J

=(141312 / 321)4=(14132)4=360

09%8hat is the sum of all digit numbers formed using the digits 2, 3, and & without

repetitionG

'planation H

!eference " .um of all num*ers formed from iven diits%

If all the possible n digit numbers using the n distinct digits are formed, the sum of all

the numbers so formed is e4ual to

(n%1)# (0um of the n digits) (111 ... n times)

ere n$.

ence the sum of all digit numbers formed using the digits 2, 3, and & without

repetition

$ (%1)# (2 + 3 + + &)(1111) $ 3# 1 1111 $ * 1 1111 $


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'planation H

!eference " #ore /seful !elations - Com*inations%

If n5'$n5ythen either ' $ y or (n%') $ y

n5;$n52:

$> n C ; $ 2:

$> n $2: + ; $ 3&

8.%6ind the number of triangles which can be drawn out of n gi!en points on a circleG

'planation H

!eference " Num*er of trianles formed *+ ,oinin n points out of which no three points

are collinear%

5onsider there be n points in a plane out of which no three points are collinear. "henumber of triangles that can be formed by /oining these n points are

n(n1)(n2) / 6

0ince all these m points are on a circle, no three points are collinear.

ence the re4uired number of triangles $n(n1)(n2) / 6$ n5380%In how many ways can 1 boos be arranged on a shelf such that a particular pair of boos

should always be togetherG

'planation H

8e ha!e a total of 1 boos.

Bi!en that a particular pair of boos should always be together. ence, /ust tie these

two boos together and consider as a single boo.

ence we can tae total number of boos as


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'planation H

"otal number of ways in which we can arrange 1 boos on a shelf

$ 11$ 1# %%%%%%%(-)

=ow we will find out the total number of ways in which 1 boos can be arrangedon a shelf such that a particular pair of boos will always be together.

8e ha!e a total of 1 boos. If a particular pair of boos need always be together,

/ust tie these two boos together and consider as a single boo.

ence we can tae total number of boos as


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ence, this must be the answer.

Nou can refer the section KnumbersKfor more information on this)

8


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ending ith 8

we tae the digit & and fi' it at the unit place. "here is only 1 way of doing this.

1

0ince the number & is placed at unit place, we ha!e now < digits(,1,2,3,,*,:,;,


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0ince one digit is placed hundredth place and another digit is placed at unit place,

we ha!e digits remaining. -ny of these digits can be placed at tenth place.

& &

"otal number of 3 digit numbers using the digits (3, , &, , *, :) $ & & $ 1

ence, re4uired number $ 1

>/%ow many & digit numbers can be formed using the digits 1,2,3,,D < such that no two

consecuti!e digits are the sameG

'planation H

ere, no two consecuti!e digits can be the same

"he ten thousands place can be filled by any of the < digits (1,2,3,,D


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Adv. Algebra

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