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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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Chapter 1 Permutations and Combinations
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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Chapter 1 Permutations and Combinations
'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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Chapter 1 Permutations and Combinations
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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Chapter 1 Permutations and Combinations
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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Chapter 1 Permutations and Combinations
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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Chapter 1 Permutations and Combinations
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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Chapter 1 Permutations and Combinations
'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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Chapter 1 Permutations and Combinations
+ 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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