Woder

mattersand
(permutations
4or16possible
222 selections
witlh repeMiAiomen:s ()
How 223

wy xy YY Z2
yx
zy Mathematmanyics ?
Compule the tollowing:students are taking Physics but not
3
yz
9!
but repetition is not allowed, 6
Horder matters
8!
(i)

without reherpelei oarne4

possibilities (permutations (39) ! 48 (5)
times 3or12 (M) (38) | (n+ 1)!
wy xy Feyr pers ns enter a
(n-1)!
yz zy 4. first
which there are six seats. In class railway
how many Compartment in
order doesnt matter but their seats ? List the ways can they take
( repetitions are various
are 10 possbilities (combinations with
repetialtiloon)wed,: there 5 How many natural
with the four digits 2,numbers
arrangement s. formed
of four digits can be
3, 5, 7, no
digit being used more than
once in each number ? What will be their number it the
aiven digits are 2, 3,5,0 ? Write allthe
numbers so formed.
6. How many natural numbers, each having
(M) Finally it order doesn't matter and digits, can be formed with the digits 0, 2, 3, 5,three ditferent
8, 9, it
allowed, there are 6possibilities repetitions aren't none of the digits is repeated ?
repetition): (combinations without (0) the digits can be repeated ?
yz
7. In how many ways can a cricket eleven
wy xz choose a
captain, a
WZ. vice-captain and a wicket-keeper from amongst them
selves?

EXERCISE 9.1 8. There are 5 competitors for a prize in high jump, 8

for a
prize in 100 metres race, and 7 for a prize in
1. If A={0, 1,2,3), B (2,3, 4)
and C= (4, 5, 6, 7, 8), find howmarny different ways it may be pOssible slow-cycling. In
to award these
the following: prizes ?
(0 O(AUB) 9
() O(ANB) A gentleman has 6 spare rooms for guests.
()
O(A)+0 (C) (v) o (AUG) ways can he In how many
acCommodate 3 guests, each in a separate
O(Ax) (vi) O(AU(BxC)) room ?
(vi) o((AUB)*(AUC))
2.
(vii) o(AU(BNC))· 10. In a class there are 8 boys and 5 girls. Two
There are 53 students taking tatives are to be chosen. In how many ways class represen
both, 38 students taking Mathematics or Physics selected it
they can be
taking Physics. Mathematics, and 40 students 0 both are chosen fromall ?
(0 How many students are taking both Mathematics and () one is to be aboy and the other a gid ?
Physics ? () the first is to be a boy and the second any of the boy
() How many students are taking Mathematics but not or gid ?
Physics ?
 questions O

Br the
6
can
nne-fakse questions
6
be answered ? how many The 225
may2fair dice oome up ?
Thore ways
diterent ways help of the utier
diierert

fol owinlieoperremnutSalions
11.
formed out can be determingd with the
many of the Theorem 9.4,
12

13.
n
how

wordmay
How COURAGE?
wordscan

them
be

wi begin with R?
o \el ers he The number
at a time is given by
of
of n elerments of a set taken r
Howmary ofthem
many of begin with Aand end wih E?
How - n(n -
vowels 1) (n- 2)
In howmany
ofthemthe Come logether ? (n-r 1), upto r factors
at Hyderabad, where (n-r)!
thethere are 6
arive
students ways can they join Proof: Let S =(4,.,
14. Four how many
olleges. In
eachata
diierentcollege? colleges, number of permutations of n
set arethe ordered
,}be a set of n

r-tuples
elements elernents. The
taken r at a time irom this
which can be forrned by
dififerent natural numbers can be
15. How may
the digits 0, 2,
3, 4,5, 6lying between10 and
1000 ? formed using elements. The number of
permutations
of r
the same as the number of ways of filling up elements
r blank
using its
out of n is
16. How many diferent natural numbers of 3 ditferent digts ordered r-tuple. places in an
each can be formed from the digits 1,2, 3, 4,5,7, il each The first element of the
even? ordered rn-tuple can be chosen in
umber is to be () odd, () Vs because there aren
elements of S for us to choose irom.
After the first element of the r-tuple is
The leters of the word POST are arranged in chosen in any cne of
17.
order. What is the rank of the word STOP" in this order ? dictionary these n ways, we are left with (n-1)
b cecond element of the elements of the set S. Thus
r-tuple can be chosen in (n - 1)
The leters "CFOSU are anranged in dictionary order. Whal ways. Therefore, by the Product Principle, the
18. fihe r-tuple can be first two elements
isthe rank of the word "FOCUS" in this order ? chosen in n (n -1) ways. After the first
MO elements of the r-tUie are
19. How many three-element subsets are there of the five. nln -1) ways, we are left wi!h (n -2)
chosen in any one of the
the third element of r-tupig can beelements the set S. Thus
element set (1, 2, 3, 4, 5)? Enumerate them. of
Again, by the Fundamental Principle, the chosen in (n -2) ways.
20. How many four-elemert subsets does the set {a,b,cd) the r-tuple can be chosen in n(n - 1) (n first three elements of
have ? - 2) ways and so on.
Continuing in this way, tiie number of ways in which the r
|9.4 NUMBER OF PERMUTATIONS elements of the ordered r - tuple can be chosen is

In many problems concerning permutations, we are not n (n- 1) (n-2) ... upto r factors.
always interested in the acual permutations formed. We only
want to krnow the total number of all the possible permutations Therefore, the number of possible ordered r-tuples is
that could be forned without actualy isting them. = n(n-1) (2-2)... [n-
numberof
(r-1)]
The symbol "P,,rSn, is usedto denote the = n(n-1) (n-2)... [n -r+1]
pemutations of n distind objeds taken r at a time. |n (n-1)(r-2)...n -r+1)](n-r) (n - r
-1)...3.2.1
Note: Some authors use the symbols nPr, P ., P(n,r)etc. n!
(n-r) (n -r-1)...3.2.1
insteadof P,. HenceP, (n-r)!
 Q r e .h r n ,Ihen
28 n! 227

(n-n)!
n! =n!
tions. Sometitneg
The metlvod emutations with some restric-
(he following examples.
then forrnutations is explained by
r=0 , n! EXBImple 4. |How
Go 2 1. ary
Ioislinsiaials
n!
diterent colours, by
n!
above the other ? gi or can2 or3be made ith Ive flags of
0r 4 or all of them, one
digit can be
Assuming that no whole Solutlon: When only 1 flag is to be
Ex8mpie 1.
three digit
mary diterent
positive repealbeed,,how
numbers can number of signals displayed at a time, the

with thenumbers
2, 3, 7, 8?
5, lormed P, -
5!
4!
=5
permutations of 5 When 2 flags are to be
Solution: Thisisthe case of digits 3a
numbers that can1aki displayed at
be ng
a tirme, the number of
total of three-digit signals= "P, = 20.
atime. Hence the
lormed 3!
When 3, 4 and 5 lags are to be displayed at a time
5! 5! 5.4.3 (2 !9) 5,
= 60. he number of signals are P,, P. and 'P., respectively, 1
=
(5-3)! 2! 2!
5! and 5! ie., 60, 120 and 120.
Example 2. How many diferent arrangements can be made by 2! ' 1! 0!
"EQUATION" ?
using all the letters of the word Hence the total number of signals
Solution: There are 8 different letters in the Word =
P,+P, +P, +P, +P,
"EQUATION" and all the 8 letters are to be used. This is a caeot 5+20 + 60 + 120 + 120
permutations of S diferent things taking all of them at a time 325.
Hence the total number of arrangements is
Example 5. In how many ways can 5 books on Chemistry and
P, = 8! = 40320. 4books on Physics be placed on a shelf so that the books on the
same subject always remain together ?
Example 3. I "P,=12 2P,, find n
Solutlon: The 5 b0oks onChemistry can be arranged in "P,
Solution: *P,= 12 P, ie., 5! ways among themselves.
n(a -1) (n -2) =12. (5) (-1) -2) The 4 b0oks on Physics can be arranged in P, ,ie. 4!
2 (n -1) (r -2) =3 (n -2) (n -4) ways among themselves.
2 (n'-3n +2) = The 2 groupings of the books on the two subjects can be
or 2n'-6n+ 4 =
3(n'-6n + 8)
3n'-18n +24 permutated in P, ,i.e., 2! ways.
or
-12n +20 = 0
or n = 10, 2. Hence the total number of arrangements
= 5!.4!.2!
But n =2
is inadmissible,since n 23. = (120) (24) (2)
Hence n= 10. 2 = 5760.
 vacant of
ditterent posts
undergraduates, whicth 3 mus
There
are 7
and 2
by
graduates or while Ihe original one
2 29
228
ABnole6

be
emaining2
graduates

and
givento either
hekd may be 5 undergraduates
by
posts be
filled2
applyfor undergraduat
these posts,esin
Ilence
get(r IP) permulations.
oul of the
Poriginal perrnutations,we shal1
gratuales
the Similarly, il we now
ways can
Aw
W10
mary and
graduates 3 different
posts tor things, different
permute them in all
from eachreplace the s like things by s unlike
other and from all the rest, and
There are
10
therefore, be filled in p. pOSsible
ways
permutationS out arrong
themselves, we obtain
Solutlon:
alone. These
can, ways. P)
(r! s! obtained.
previously of ihe (r! P) permutations
graduates undergraduates can fillthe 2 posts for them
Similarty. the 5 Qeneat the process with the t like things of the
akone in P,
ways.
posts still remain to be filed replacing them by t un!ike things different from eachthird
otherkind,
and
has been done, 2 rest, and
When this candidates for the same. Thus the from allthe permuting them all possible ways amon9
in
remain 10 the themselves. The number of permutations is thus increased t!
up and there
remaining 2 posts
can be filled up by remaining 10
times. Hence we
have(r!: s! t! P)permutations in the end
ways. things are now all different, and the total number ot
candidates in "P But the n
number of ways required to fill all the their permutations must
be n !
Hence the total
5! 10! Therefore r!'s! t! P=n!
posts 10!
n!
-"P,.P.. P,= 7! 3! 8!
or P= r! s! t!
(9)
(10) (9) (8) (5) (4) (10)
1296000. This number of permutations is symbolically written as
permutations of
restricted ourselves to
We have so far
things which were all ditferent. We now proceed to consider tha
method of
(.)
some of the things may be alike. The
cases in which
permutations when some or all of the The above theorem can be easily extended to cases
finding the number of folowing theorem. where there are more than three kinds of like things.
things are alike is given in the
Note: If r,s and t are such that none of the things are left
Theorem 9.4.2
pemutations of n things taken all at Over, then r +s +t=n.
The number of distinct alike cf
alike of one kind, s of them Example 7. Find the number of permutations of 10 balls.
2 time, when r cf them are fany. al
the rest,
another kind, t ot them alke of a third kind and taken 10 at a time when 3 are black, 4 are red and 2 are whte.
Solution: Here 3 black balls are like things of the tirst kind, 4
Cterent is
balls of the
red balls are like things of the second kind and 2 white required
balls is 10. Hence the
ifo kind. The total number of
Proof: Let P be the reuired number of permutations.
Take umber of permutations is
any one of these permutarrs and replace the r like things by r
and 10!
unike things. citferent frcm each Cther and from at the rest, 10
pemute themiin al posst' was among thhemselves, leav9 3! 4! 2!
of:ne
3,4,2 10.9.8.7.6 5.(4 !)
e cthers unchaged. ie'tri
permutations Out 3.2.1.(4 9.2.1
12600
 ayS an the 31
Solon.
evired The 5
pemselves
31
Rs are tog 21
e e sin the word ARRA,GE The odd dgts3
e rema nng 3 are all c'ergr place3 i
sevesin ttheir four
3 rpetons s 41
= 6 waf
2! 21

2! 2! Cach of the ways in ( 2 A 2 e n s e

Hence the required
76543.2 waYS (m).
in
=
(2.21 Erample 10. A coin is tossed seven tines
ray possible ways can we get 4 heads ard 31als.
= 1260.
Solutlon: Out of 7tosses, we are to have 4 heads and 3tails
treat tnem as one laz.. done in
R5 2: together. we can
N. G. EE. They can be This can be
ieters A RA, A 7! 7.6.5. (4 )
"e we r2ve6
of permutations in.. 4!3! (4!).3.2.1
- : 353 wavs. Hence the number
2! therefore the n 35 ways.
e wo Rs are 13gether s 350, and
are never togeher:
eUtatcns in wich they
i EXERCISE 9.2
283-35 =901.
AA as on:
As the tWO AS are to be t0gether, we can treat Find the value o!:
all. They ca
ieer so ta we have AA, R, R, N, G, E, Six letters in P, () (m)
6! = 360 ways. If, n
De pemted among themselves in 2!
6 \
addtion, the two Rs are also to be tOgether, the number of suc:
pemutations of AA, RR, N,G, E will be 51 = 120. Hence ths
(M) (6:) 2. 198/

required number of permutations in which the tWO A's are 2. Find n, it

together but not the two R's is 360 - 120 = 240.
*P = 20 (i) P,= 2.(*P,)
the two A's are together but not the two R's, we have 240 (0 2n -'p = 3:5.
pemutaions. Sirrilarty , t two R's are together but not the two
As, we have 240 permutations. Also, if the two A's as well as trie
two R's are together, we have 5!'= 120 permutations. Hence ti From a board of 18 members, one director and one deputy
required number of permutations in which neither the ASI Oirector is to be chosen. In how many ways they can be
the Rs are together is selected ?
1260-240-240 -120 = 660. examination, two of which
Oevenpapers are to be set at an
Example 9. How many positve whole numbers can beformed many ways can the
dre on Algebra and Calculus. In how
papers are not
with the digts 3, 4, 5,6.
5,4,3 so that the even OCCupy the Papers be set so that these two
even places ? digits Consecutive?
 friendly malcheS, each Findthe
232
DlayA0ainstevery
cricket
teams
play
one oftherest How many matcheshav
haung the following
THATA
words al
1aken
233

letters of each of
5
Eioht
in
plaYed
al2
shirtsand 8pairs of
(D PAKIS IAN (n)
iongttor
o be
coats, 7
has 6
AmanwAyscanhe
make up a complete su oftrouserclosth.esIn ho
2 15.
Four out of 10
are of dilerent
balls are rod NSTTUONS
colOUrS of ballsorme canaro hegeen. and thein rest
ways, find the nurmber MIhe
shirtsin 12 choices of
6 many
manutacturer
makes
lengths. How
in 33sleeve must
and 1abrmanyicsdi,tferer
each How many natural n
green balk, arranged 6300
A
in8
neck
shinstrom
s0zeS manutacturer
this
a store carry to
a have
16.
nurnbers
of the 5 digits 1,2, 3, 4, 5 may be
tormer
by using 4 Out
complete stock? it the digits are not
diferent signals which
numberof and 2can be made
t the digis may be
# repeaterd 2repeated
in ary rumter ?
Findthe which are white, 2 black red,
all the 6 ( thedignts are rot
repeated and
2 of becin with
wth6flags,usedevery time.
fiagsbeing
Kthe digis are rot
repeated and end wth 252?2
77 different books be In how many ways can 3 English, 2 Urdu and 2
9. In how
In
many wayscan
how many ofthese ways
arranged
will a specified
book
on a
be
17.
hooks be arranged on a shet so as to keep al the bocks Sinchi
n
shet? () at either end ? each language together ?
centre ?
at the
arangements can be made by Fnd the number ofways in which 5 fist year students and
How mary ditferent "MATHEMATICS" ? How using al
18.
10. word 5second year studerts can be seated on a bench in an
beginofwith C* ? How many of them begin withmany
the w of
themletters
the "T"?In examination so that no tWO studets of the same cdass are
will OCcur together ? next toeach other.
how many of them consonants
19. Three prizes are offered to a class of 50 students. In how
formed from h.
11. 0 How many different words can be many ways can they be given,
"DAUGHTER" ?
letters of the word it they are for proficiency in three ditterent subjects.
in how many of these words vowels OCcur together 2 0 they are the first, second and third prizes for
() Howmany of these willbegin with D ? proficiency in the same subject ?
(v) How many of these will begin with D or T?
(M In how many of these the letter R will occur at the 20. Find the total number of ways in which 5 sparows can
fifth place ? perch on 3trees,when there is no restriction to the choice
() HoW many words can be formed if vowels are to be of a tree. In how many of these ways wil one particular
placed at second, fourth and sixth places ? sparrow be alone on a tree ?
12. How many diferent natural numbers can be formed ym 9.5 CIRCULAR (OR CYCLIC) PERMUTATIONS
between 1000and 10000 out of the digits 0, 1,2, 5. 7,9!
How many of them wil neither be divisitble by 5 nor by 21 Now we shalldeal wth circular pemutations. In an ordinary
13. (0 How many three-digit natural numbers Can be permutation of r things, the elements are aranged in a definite
VOer and their places can be maked as the first, the second, the
formed fromthe digits 0, 1,2, 3, 4,5 and 6 each d¡l elc., so that every ararngement has a beginning and an
can be used only once ?
How many of these are odd
third...
end. In a circular permutalion, wherethe elements are arranged
(M) HOw many are numbers ? rOund the circumterence of a cice, there is nether a beginning
greater than 330 ? nor an end, and the positions can n marked oulabsoiuiely
 the
nmeratorand denominator of (he
the
Ag
right prample1.
Find Ihe value of 239
( n .+ solutlon:
B(-1}
(n-2)... 1)[(n ))
rl(n-r)! 101
() (10 4) ! 4!

10987-6!
6!4321
: r=R,then
o 210.
()- 1,
ie. the uTber of combinations of nthings taken n (all) at atime
Erample 2. "C. ,find Cs
solutlon: Here
orley2( ) ().
theorem, we have
Then by Theorem 9.6.1,
From the above
18! 18!
n!
(18-r) !:r! (18-r-2) ! (r +2) !
(} n-(a-r)]!2-r)!
n! Cancelling out 18!:
r'in-r)!
(18-r)!r! (18-r-2) ! (r +2)!
=
() 1 1
When we are selecting r elements Out of n, in fact, we are ie.
(18-r)lrl (18-r - 2)!}]· +2)(+1)·r!
iezving -r)elements. Sometimes it is simpler to select (n-r)
(n
elementsS which are left out. For example, if we have to select 90
i.e.
studets ou of 100, t would be simpler to select 10 students (18-r) (18-r-1)[(18-r-2) !]·r!
whom we want to reject. Such combinations of r elements
or. in - r jelements, out ofn elements are called complemen
tary combinations. [(18-r-2) !]·(r +2) (r +1)·r!
vdnceling out (18-r-2)!]r! from the denominators
Corolary 3 ()--0)-() - 1. 1
1

In terms of sets, t means that the number of null sets in a (18-r) (18r-1) (r+2) (r + 1)
set of n elements is 1. Alkernatively, it means that there is only
one way not to t§ke any element from a set
ofn elements.
Cross-multiplying:
r+2) (r + 1) = (18-r)
(18 - r-1),
(17-r)
Also,
(1)-a ie.
i.
(r+2) (r+ 1) = (18-r) +r
P+3r+2 =
306-35 r
 teathers
Erample tions, Ths comparison,
Therefore by Bt
was
in}ways
eachers Solution: or or
we AlematSo, ively.,
Sometimesas are
Hence,by be 3. is given
fomed seen 18-r=r2 we
The n that know
in
2730. =
-(9)6)
(15-3)!-3!
15-14-13-121 the studerts out
how
the we
maryfollowingcome before.
56,as =
f= 8 16
2r=
=
thal 56.
=
8. 304
12!-3-2-1 15! Produc
15
of "c (8-5)!(59
3! 8-7-659
8-7-6
can ways =
studerts acrosS
Principle, be can example.
(4-)1-2! chosen combinations
4! andparty a 32187:6
4-3-2!
2!-2! the
in( teachers 4of
required 15
3
)ways students
with
?
number
andthe rectis
and2
of

(ü)
done in are
This other solution: Exaoumt ple
players14
(@) 0
of

to
Ifaparticular Hence canplayers
partcular a 4.
Hence select be incue
isexcde
14
The \no

11 (10) the done are () number How

?

() the ways.
required
players
player
required player
be
to 364. =
(14-11):111
1211!
1413
a mary
paiialarmary
pañiaNar
a

78. =
of
11!13
12 12
13
11 (13-10)
13 10! ! selected l5 311! 14|
ways
ays
of
2!11!· (13-11)!1 ! out is 32:1 number
13!
ways.
t0 player?pdayorihem
10!3! 12 a
13! number of to oe ol
the be 11 always
irom selecing ? a
always
total of aleven
cket
of 286. = 10! combinations the
13-12 combinations of included, 14-13-12
21 excluded,
13 renaining 11
321
players. players
be
then chosen
then is
is This 13 in out 241
in players. fact
can fact ot
10
bewe
 () 3
N

is
frequently called PASCAL'S RULE These Can
(M)
be
profes 0r3 nthdort,
Aprofessors arn 1
chosen 0
243

9.6.1, We have
isingTheorem
n!
() )
n/

(n -r+1) (r -1)! (n -r)!r!

(i)
ways respectively.
() ) 4

n!r+n! (n-r+ 1) All these choices

(n-r+1) !r! adding traclions Principle,the required are
nurmber inofdependent
ways is Hence by the Sum
n! (r+n -r+1) ,factorising the numeralor
(n-r+1) !r!
4 x 56 +6 x28 + 4 x 8 4 1

-("*). using Theorem 9.6.1 again = 425.

EXERCISE 9.4
Example 5. Show that . 6)-()-(0). Find the value of:
1.
Solution:
8! 8! () (i) (M)
6)- ( (8-3)!3!
8! 8
(8 4)! 4!
8! 8!
2
Find n,it
"C, = 44:3,
(0
5!3! 4!4! 54!3! 4!43!
(i) 220
8!4+8!5 (4 +5)8! (i) P, = 24 "C.
5!4! 5!4!

9 8! 9!
3 M
().
5!4! (9-4)!4! () (6) - ind ().
4. Find nand r,if
"P, = 240 and "C, = 120
Example 6. Adepartment in acollege consists of 4 protessors (0 "P, = "P.., and "C, =nC,
and 8students. Astudy tour is to be arranged. In how many ways
can a party of four be chosen so as to include at least one 5 Find the number of combinations of the lettersof the word
professor ? "SEMINAR", taken 4 at a time.
Solutlon: The party may consist of 6
How many triangles are determined by 12 points in a plane,
1professor and 3 students, or I the points are used as vertices a
(ü) 2 prolesSors and 2 straight line ?
students, or 0) io 3points lie in the same
 samestraight line 2
lle nthe
in
oints () at least A
commitleo
"o#e
and4
ladies,
can
a
this be done Out of 12 persons oiers and at leasi 245
6etlemen
many
ways
15 how maty one iet
parlyto) Consist of not
From how
In one
lady?

There are 10balls

rrore thanpar57ties can he
lorned lcast
female
cOuncillors in a formed 2c7
canone select 7 of
At
inne

ConSiStin
Thereare
20
how"
male
of5
andA
ways a
male
many and
12
committee
temale
can
councillors? \nbehoiManymuniloCr pamelg 16
o include thethern
to exclude the
sO
red ball as,
white ard
29mmittees
In of7therewillbe at most 2female CouncilOI5) the ar
N

formed by 2
In how many diferent ways car
orie irvte
nalr, ?

How
many
words
of 4
can be
vowels and 7 consonants ? Vowels and 3 17. exactly 3 people,
3 or more people,
consonantsout 5
candidates are to
be from a group of friends to
election 3
voter can
vote utmost
as electedmanyoutas of 5 7 members is
Apartyof 5 gents. In to be
dinner?
In an
A to be elected. the 18.
how rnany chosen irom a group ot 5
candidates. candidates
number of in which a voter can
nurmberof
ways
vote ? Determine the ladies and
formed it it is to contain
exactly 4 ladies,
ways can the pary ce
mathematics there are two
In a paper
of
questions.In
quUestions. Every examinee seclhasions, each
5 many ways he can answer,if heto cannol
of how do 6
at least 4 ladies,
at most 4 ladies ?
consisting
than.4 from one
section ? father has 8 children.
Af He takes them 3 at
a
attempt more
19.
as often as he can without laking the same 3time to the zoo
of 52cards
2 cards are to be drawn. n
how than once. How often wil he go, and how children more
often will each
deck
12
From a
thiscan bedone. if child go ?
many ways simultaneously 2
the cards are drawn 20. In how many ways can 16 rupees be divided among 4
draWn one by one ?
two cards are drawn one by one but the first card is beggars SO that no beggar shal! receive less than 3
two cards are deck before the second draw? rupees?
placed backinthesimutaneoUsly drawn cards one is to
Out of thetwo o7 DIVISION INTO SECTIONS OR PARCELS
(M) be red and the other black ?
and the other is an ace Theorem 9.7.1
one of them is black drawn one by one after
() assuming that the cards are The number of ways of partitioning a set consisting of
replacement ? Ir+) elements into pairs of two disjoint subsets such that one
drawn. H sIbset consists of r elements and the other of s elements is
balls, five balls are to be
bag containing 50 black,then find
13. From a there are 5 white balls and rest contain at least (r+s)!
in the bag balls r!s!
so that the drawn
the number of ways
one white ball. Proof: Let Aand B be two disjoint sets
cricketers of whom consisting of r and s
be chosen out of 15 ements respectively. The total number of elements in AUB
11 is to keepers. In ng
14. A team of others are wicket 6r+s.
bowlers and 2
5 are team contains
that the
done so
ways can this be keeper ? NOW the total number of all such
many
4 bowlers and 1wicket subsets of AUB which
(0 exactly MSsl ol r elements is
simplyC