Tut 28 POYO & HQ & Supplementary Qns (Solutions)

7/25/2019 Tut 28 POYO & HQ & Supplementary Qns (Solutions)

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Chapter 1 Permutations and Combinations

Tutorial 28: Permutations and Combinations

Practise on your own

1 A delegation of 3 girls and 2 boys is to be selected from a class of 18 girls and 12

boys. Find the number of possible delegations.

Solution:

Number of possible delegations = 18 12 38!3 2

=

2 "n a set of 2! cards# each card is mar$ed %ith one of the letters AtoZso that each

card carries a different letter of the alphabet. &hree of these cards are dra%n at

random. Find the number of different selections that can be made

'i( if the cards are dra%n %ithout replacement and the order in %hich the

cards are dra%n is disregarded#

'ii( if the cards are dra%n %ith replacement and the order in %hich the cards

are dra%n is ta$en into account.

Solution:

'i( Number of different selections =2!

2!))3

=

'ii( Number of different selections = 32! 1**!=

3 A nursery school has + apples# 3 oranges and 2 bananas to share among , children#%ith each child recei-ing one fruit. Find the number of different %ays in %hich this

can be done.

Solution:

Number of %ays =,

12!)+32

=

+ A child %as gi-en four bo/es of toys. "n the first bo/# there %ere three identical toy

cars. "n the second bo/# there %ere four identical toy -ans. "n the third bo/# there

%ere t%o identical toy motorcycles. "n the last bo/# there %as a toy garbage truc$.Find the number of %ays in %hich the child can choose at least one toy from any of

these bo/es.

Solution:

3Cars

'+%ays(

+0ans

' %ays(

2&Cs

'3%ays(

1&ruc$

'2%ays(

Number of %ays child can choose at least one toy

= ( ) + 3 2 1 child chooses no toy = 11,

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A set of 2) students is made up of 1) students from each of t%o different year

groups. Fi-e students are to be selected from the set# and the order of selection isunimportant. Find

'i( the total number of possible selections#

'ii( the number of selections in %hich there are at least t%o students from each of

the t%o yeargroups.

Solution:

'i( &otal number of possible selections =2)

1)+

=

'ii( Number of selections =1) 1)

23 2

or 1)8))=

! Find the number of +letter code%ords that can be made from the letters of the %ord

A0ANC4#

'i( using neither of the 5A6s#

'ii( using both of the 5A6s.Solution: 2A# 1# 10# 1N# 1C# 14

'i( Number of +letter code%ords =

+P = 12)

'ii( Number of +letter code%ords = +.

2 2.

= 12)

* 'a( 4ight people go to the theatre and sit in a particular group of eight ad7acent

reser-ed seats in the front ro%. &hree of the eight belong to one family and sit

together.

'i( "f the other fi-e people do not mind %here they sit# find the number ofpossible seating arrangements for all eight people.

'ii( "f the other people do not mind %here they sit# e/cept that t%o of

them refuse to sit together# find the number of possible seatingarrangements for all 8 people.

'b( &he salad bar at a restaurant has ! separate bo%ls containing lettuce#

tomatoes# cucumber# radishes# spring onions and beetroot respecti-ely. ohndecides to -isit the salad bar and ma$e a selection. At each bo%l# he can

choose to ta$e some of the contents or not.

'i( Assuming that ohn ta$es some of the contents from at least one bo%l#find ho% many different selections he can ma$e.

'ii( ohn decides he is going to ha-e + salad items# and one of them %ill

be tomatoes. 9o% many different selections can he ma$e:

Solution:

'a( 'i( Number of possible arrangements = ! ; 3 = +32)

'ii( Number of possible arrangements = +32) < ; 3 ; 2 = 288)

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'b( 'i( Number of different selections = !2 1 = !3

'ii( Number of different selections =

1)3

=

8 4ight cards each ha-e a single digit %ritten on them. &he digits are 2# 2# +# # *# *# *#* respecti-ely.

Find the number of different *digit numbers that can be formed by placing se-en of

the cards side by side.

Solution:

Case 1 4ither digit + or are not included

Number of different *digit numbers =*

22+

= 21)

Case 2 >ne of the digit 2 is not included

Number of different *digit numbers =*

+ = 21)

Case 3 >ne of the digit * is not included = * +2)23

=

&herefore# total number of different *digit numbers = 21) ? 21) ? +2) = 8+)

Challenging Questions

1. 9o% many rectangles are there in this figure: 9o% many rectangles are therein an

m

/n

grid: @1)#

' 1( ' 1(

+

m m n n+ +

Solution:

Number of rectangles = !

1)2 2

=

Number of rectangles in an m; ngrid = 1 1 ' 1( ' 1(2 2 ' 1('2( ' 1('2(

m n m n

m n+ + + +=

' 1( ' 1(

+

m m n n+ +=

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2. &he follo%ing diagram sho%s 12 distinct points a1# a2# a3# b1#B# b+# c1#B# cchosen

from the sides of ABC.

'i( 9o% many line segments are there 7oining any t%o of the points on

different sides:

'ii( 9o% many triangles can be formed: 9o% many uadrilaterals can be formedfrom these points:

Solution

'i( Number of line segments = ; + ? 3 ; + ? 3 ; = +*

'ii( Case 1 2 points from AD

Number of triangles =

*2

= *)

Case 2 2 points from AC

Number of triangles =+

82

= +8

Case 3 2 points from DC

Number of triangles =3

,2

= 2*

Case + 1 point from each side

Number of triangles = + 3

!)1 1 1

=

&herefore# total number of triangles = *) ? 8 ? 2* ? !) = 2)

'iii( Case 1 2 points on t%o sides

Number of uadrilaterals = 3 + 3 +

1)82 2 2 2 2 2 + + =

Case 2 2 points on each side and 1 point each on 2 sides

Number of uadrilaterals = + 3

3 + 3 + 2*)2 2 2

+ + =

&otal number of uadrilaterals = 1)8 ? 2*) = 3*8

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A

B C

c5

a3

b3

b2

c4

c3

c2

c1

a2

a1

b1

b4


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3. Code numbers# each containing three digits# are to be formed from the nine digits 1# 2#

3# ...#, . "n any number no particular digit may occur more than once.

'i( 9o% many different code numbers may be found# and in ho% many of these%ill , be one of the three digits selected:

'ii( "n ho% many numbers %ill the three digits occur in their natural order 'ie the

digits being in ascending order of magnitude reading from left to right# eg

3,(:

Solution:

'i( Number of different code numbers =,

3P= )+

Number of reuired code numbers =8

3. 1!82

=

'ii( Number of %ays =,

8+3

=

+. 'i( nred counters and mgreen counters are to be placed in a straight line.

Find the number of different arrangements.

'ii( A to%n has nstreets running from south to north and mstreets running from

%est to east. A man %ishes to go from the e/treme Eouth%est intersection to

the e/treme Northeast intersection# al%ays mo-ing either north or east alongone of the streets. Find the number of different routes he can ta$e.

Solution:

'i( Number of different arrangements =( ) .

. .

n m

n m

+

'ii( Number of different routes =

( )

( ) ( )

( )

( ) ( )

1 1 . 2 .

1 . 1 . 1 . 1 .

n m m n

n m n m

+ +

= . Find the number of %ays the positi-e integerNcan be e/pressed as a sum of integers#

each of %hich is 1 or 2# %hen'i( Nis e-en and 'ii( Nis odd. '8,CEp(

@&he order in %hich the 1s and 2s appear does not matter. e.g. 1 ? 1 ? 2 is the same as

2 ? 1 ? 1 and 2 ? 1 ? 1

Solution:

'i( Number of %ays = 12

N +

'ii( Number of %ays =

1

12

N+

!. 9o% many diagonals can be dra%n in a pentagon: Generalise the result in the case of

a nsided polygon.Number of diagonals in a pentagon =A nsided polygon has n-ertices# each -erte/ has 'n< 1( diagonal lines.

&herefore# there are( )3

2

n ndiagonals. >r

2

nn

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TEMASEK !"#$% C$&&E'E

Su((lementary Questions on Permutation ) Combination *$n +our $wn,

'Eolutions %ill be uploaded on atri/ 2.(

1. 9o% many numbers bet%een 1) and 3)) can be made from the digits 1# 2# 3# if

'i( each digit may be used only once# 'ii( each digit may be used more than once :

2. 9o% many combinations of three letters ta$en from the letters A# A# D# D# C# C# are

there:

3. A mi/ed team of ten players is chosen from a class of thirty# eighteen of %hom are boys

and t%el-e of %hom are girls.

9o% many %ays can this be done if the team has fi-e boys and fi-e girls:

+. Find the number of %ays in %hich t%el-e children can be di-ided into t%o groups of si/

if t%o particular boys must be in different groups.

. 9o% many of the permutations of the letters of the %ord A&94A&"CE do all the

consonants come together:

!. A bridge team of four is chosen from si/ married couples to represent a club at a match.

"f a husband and %ife cannot both be in the team# ho% many %ays can the team be

formed:

*. &%o sets of boo$s contain fi-e no-els and three reference boo$s respecti-ely. 9o% many

%ays can the boo$s be arranged on a shelf if the no-els and reference boo$s are not

mi/ed up:

8. A bo/ contains ten bric$s# identical e/cept for colour. &hree bric$s are red# t%o are

%hite# t%o are yello%# t%o are blue and one is blac$. 9o% many %ays can three bric$sbe

'a( ta$en from the bo/ 'b( arranged in a ro% :

,. 9o% many of the arrangements in a ro% of all ten bric$s in Huestion , are

'i( the three red bric$s separated from each other#

'ii( 7ust t%o of the red bric$s ne/t to each other:

1). "n a multiplechoice uestion there is one correct ans%er and four %rong ans%ers to eachuestion. For t%o such uestions# ho% many %ays is it possible to select the %rong

ans%er to both uestions:

11. "n Huestion 1)# if a correct ans%er scores one mar$ and a %rong ans%er scores Iero# in

ans%ering three such uestions ho% many %ays is it possible to score

'i( )# 'ii( 1# 'iii( 2:

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12. A forecast is to be made of the results of fi-e football matches# each of %hich can be a

%in# a dra% or a loss for the home team. Find the number of different possible forecasts# and sho% ho% this number is di-ided

into forecasts containing )# 1# 2#3#+# errors respecti-ely. 'J of K(

13. Find in factor form the number of %ays 2) boys can be arranged in a line from right to

left so that no t%o of three particular boys %ill be standing ne/t to each other.'J of K(

1+. Find ho% many distinct numbers greater than )))

and di-isible by 3 can be formed from the digits 3# +# # ! and )# each digit being used at

most once in any number'D(

1. A certain test consists of se-en uestions# to each of %hich a candidate must gi-e one of

three possible ans%ers. According to the ans%er that he chooses# the candidate mustscore 1# 2# or 3 mar$s for each of the se-en uestions.

9o% many different %ays can a candidate score e/actly 18 mar$s in the test: 'K(

1!. A tennis club is to select a team of three pairs# each pair consisting of a man and a

%oman# for a match. &he team is to be chosen from * men and %omen. 9o% many

different %ays can the three pairs be selected: 'J of K(

1*. Eho% that there are 12! %ays in %hich 1) children can be di-ided into t%o groups of .

Find the number of %ays in %hich this can be done

'i( if the t%o youngest children must be in the same group#

'ii( if they must not be in the same group. 'J of K(

18. A committee of three people is to be chosen from four married couples.Find in ho% many %ays this committee can be chosen

'i( if all are eually eligible#

'ii( if the committee must consist of one %oman and t%o men#

'iii( if all are eually eligible e/cept that a husband and %ife cannot both ser-e on

the committee. 'J of K(

1,. Find the number of integers bet%een 1))) and +))) %hich can be formed by using

the digits 1# 2# 3# +

'i( if each digit may be used only once#

'ii( if each digit may be used more than once.

2). 9o% many different %ays can the letters of the %ord A&94A&"CE be arranged:9o% many of these arrangements %ill t%o ALs be ad7acent:

Find the number of arrangements in %hich all the -o%els come together. 'J of K(

21. Find the number of different arrangements of the letters in the %ord P4NC"KE in %hich

'i( the 4 and the "are together#

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'ii( the 4 precedes " 'not necessary 7ust precede(.

Solutions to Su((lementary Questions

1 'i( 2digit nos. 3 2 = ! # 'ii( 2digit nos. 3 3 = , 3digit nos. 2 2 1 = + # 3digit nos. 2 3 3 = 18

&otal = 1) %ays &otal = 2* %ays

2 &%o cases

Case 1 all different

3

+

= +

Case 2 1 repeated

1

3

1

3= , #

&otal = 13%ays

3

18

12

= !*88!

+&a$e the t%o particular boys out# no. of %ay to di-ide 1) boys into t%o groups =

1)

2 and there are 2 %ays to assign the t%o boys into 2 groups. Ans 2 @

1) 2

= 22

###C#EM A#4#A#"

Ans.2

.

.2.2

.* = *!))

!. 1st member 12 %ays2nd member 1) %ays

3rd member 8 %ays

+th member ! %ays

Ans .+

!81)12

= 2+)* 3 2 = 1++)

8 'i( 3 # 2 O# 2 # 2 Dlue# 1 Dlac$ .

Case 1. All the same 1 %ay Case 2- All di..erent :

3

/ 01 ways

Case 3. 2 same 1 different

1

+

1

+

= 1! %ays &otal = 1 ? 1) ? 1! = 2*

'ii( 1 ?

3

3 ?

1

+

1

+

.2

.3 = 1),

, 'i( Q Q Q Q Q Q Q 'ii(8* .

2 *)!)32.2.2.

=

red

Ans 328)3

8

.2.2.2

.*=

1) Ans + + =1!

11 'i( + + + = !+ 'ii(

1

3

+ + = +8 'iii(

2

3

+ = 12 or 1 1

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+ 3

12no. of errors

) 1 2 3 +

no. of %ays1 1)2

1

=

122

2 =

12

3

3 =

12

+

+ =

12

=

&otal = 2+3 %ays

13

Ans 1*

3

18

3 = 18 1* 1!

1+ A no. is di-isible by 3 if its sum of all digits is di-isible by 3.

Case 1. Jse +##!#) Case 2. Jse 3#+##) Case 3. Jse 3#+##! Case +. Jse 3#+##!#)

or

!

or

! 2 3 2 1 1 3 2 1 2 3 2 1 + + 3 2 1 = 12 %ays = ! %ays = 12 %ays = ,! %ays

&otal = 12 ? ! ? 12 ? ,! = 12! %ays

1 Case 1 Huestions score 3 mar$s# Case 2. + Huestions score 3 mar$s#

1 Huestion scores 2 mar$s 3 Huestions score 2 mar$s#

and 1 Huestion scores 1 mar$. ) Huestion scores 1 mar$.

Ans

*

1

2

1

1?

+

*

3

3

= **

1!

3

*

3

3 2 1 =

21))

1, 'i( 3 3 2 1 = 18

'ii( 3 + + + = 1,2

1*

1)

2 = 12! 'i(

2

2

3

8

= ! 'ii( 12! ! =

*)

18'i(

3

8

= ! 'ii(

1

+

2

+

= 2+ 'iii(

32.3

+!8=

2) +,8,!)).2.2.2

.11= # ,)*2))

.2.2

.1)= # 12),!)

.2

.+

.2.2

.8=

21 'i( ! 2 = 1++) 'ii( ! ?

2

!

= 22)

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Tut 28 POYO & HQ & Supplementary Qns (Solutions)