Permutation & Combination

In this chapter, we include only rules (ie Quicker Meth-ods) based on the questions asked in the various exams like, CAT, MAT, XLRI, FMS, Bank PO, AAO, Provident Fund, CET, UT1 etc. For basics, please refer to 'Magical Book on Quicker Maths'. Many aspirants find difficulty in under-standing the basics of "Permutation and Combination". Therefore we advise you to go through all the rules dis-cussed in the following pages and try to understand the detail method'. Still you are unable to understand, just mug

the rules, apply to the appropriate questions and get the desired answers. Since weightage of this chapter is not much, only I or 2 questions are asked in the various competitive exams mentioned above, we again advise you to stick with these rules your purpose will be served.

Some Important Notations and Formulae From the examination point of view the following few

results are useful. Without going into details you should simply remember the results. 1. Factorial Notations

The product of n consecutive positive integers be-ginning with 1 is denoted by n! or |n and read as factorial n. Thus according to the definition of |n

Jn = 1 * 2 x 3 x ... x ( n - 1) x n = n x (n - 1) x ( n - 2 ) x ... x 3 x 2 x l

For example, |6= 1 x 2 x 3 x 4 x 5 x 6 = 6 x 5 x 4 x 3 x 2 x 1 = 720

2. According to the definition of Jn (a) |n = n x ( n - 1) x ( n - 2 ) x ... x 3 x 2 x ]

= n { ( n - l ) x ( n - 2 ) x . . . x 3 x 2 x 1} = n(n - 1 ) {(n - 2 ) x ... x 3 x 2 x l } and so on.

.-. |n = n[n- 1 = n(n - 1) |n - 2 = n ( n - l ) ( n - 2 ) In - 3

(b) If r and n are positive integers and r < n, then n\ Mx(f l - l )x(r t -2)x. . .x(r + l ) x r x ( r - l ) x . . . x 3 x 2 x l r\ - l)x(r - 2)x...x3x 2x 1

= ( - l X - 2 ) . . ( r + l )

4.

5.

8.

9.

10.

(n-r) = n ( n - l ) ( n - 2 ) . . . ( n - r + 1)

n\ , Caution: , * I

For example,

= 8 x 7 x 6 x 5 = 1 6 8 0 * 1 - 1 ! 2> 4! U .

V -(n-r).

Where " Pr = number of permutations or arrange-

ments of n different things taken r at a time.

"Cr re ri (n-r)

Where "C,. = number of selections, or groups of n

different things taken r at a time. From (3) and (4), we have

"Pr = r ! x "Cr

Total number of arrangements = total no. of groups or selections * r!.

I f "Cx = "Cy then either x = y or x + y = n

Number of permutations of n things out of which P are alike and are of one type, q are alike and are of the other type, r are alike and are of another type and

remaining [n - (p + q + r)] all are different = , , ^ .

Number of selections of r things (r < n) out of n identical things is 1. Total number of selections of zero or more things from n identical things = n + 1. Total number of selections of zero or more things from

642 PRACTICE BOOK ON QUICKER MATHS

11.

n different things

= " C 0 + "Cj + " C , + . . . + "Cn=2" .'

Number of ways to distribute (or divide) n identical things among r persons where any persons may get

any no. of things =

12. (a)

(b) 0! = l

(c) " C r = " C n - r =

'C r - l

! (n-r)

(d) " r = " C n = 1

13. Some Fundamental Principles of Counting

(a) Multiplication Rule Suppose one starts his journey from place X and has

to reach place Z via a different place Y. For Y, there are three means of transport - bus, train

and aeroplane - from X. From Y, the aeroplane service is not available for Z. Only either by a bus or by a train can one reach Z from Y. Also, there is no direct bus or train servie for Z from X. We want to know the maximum possible no. of ways by which one can reach Z from X.

For each means of transport from X to Y there are two means of transport for going from Y to Z. Thus, for going from X to Z via Y there will be 2 (firstly, by bus to Y and again by bus to Z; secondly, by bus to Y and thereafter by train to Z.)

+2 (firstly, by train to Y and thereafter by bus to Z; secondly, by train to Y and thereafter again by train to Z.)

+2 (firstly by aeroplane to Y and thereafter by bus to Z, secondly by aeroplane to Y and thereafter by train to Z.) = 3 x 2 = 6 possible ways We conclude:

I f a work A can be done in m ways and another work B can be done in n ways and C is the final work which is done only when both A and B are done, then the no. of ways of doing the final work, C = m x n.

In the above example, suppose the work to reach Y from X = the work A in m i.e. 3 ways. The work to reach Z from Y = the work B in n i.e. 2 ways. Then the final work to reach Z from X == the final work C > in m x n, i.e. 3 x 2 = 6 ways.

(b) Addition Rule Suppose there are 42 men and 16 women in a party.

Each man shakes his hand only with all the men and each woman shakes her hand only with all the women. We have to find the maximum no. of handshakes that taken place at the party. Case 1: Total no. of handshakes among the group of 42 men

= 4 2 C = -42! 42! 42x41x40!

' 2 2! ( 4 2 - 2 ) 2! 40! 2 x 1 x 4 0 ! = 21 x 4 1 = 8 6 1

Case 2: Total no. of handshakes among the group of 16 women 16! 16x15x14!

2! ( 1 6 - 2 ) 2x1x14! maximum no. of handshakes = 86

= 8x15 = 120

120 = 981.

I. Permutations Rule 1

Problems based on direct application of the formula.

"P. n\

(n-r).-

Working Rule

( 0

(ii)

(iii)

nP, n\

= n(n-\fn-2)... tor factors. (n-r).

I f LHS is the product of r consecutive integers, ex-press RHS also as the product of r consecutive inte-gers. Factorise the RHS, find out the greatest factor and tr> with that factor. I f greatest factor does not suit then try with greatest factor x least factor. Look at the ex-ample given below and try to understand the working rule.

Illustrative Example

Ex:

Soln:

I f "P4 = 3 6 0 , find n.

Given "P. = 3 6 0

n\ 360 " ( - 4 )

or, ( - l X - 2 X > 7 - 3 ) = 360 = 6 x 5 x 4 x 3

.-. n = 6 [Here LHS is the product of 4 consecutive integers therefore, RHS ie 360 is to be expressed as the prod-uct of 4 consecutive integers. 360 = 2 x 2 x 2 x 3 x 3 x 5 , greatest of these factors is 5, therefore try with 5. Integers just before and after 5 are 4 and 6. Both 4 anc 6 are factors of 360. Thus we get four consecuti\integers 6, 5,4 and 3 whose product is 360. If 5 does not suit, then try with 2 x 5 i.e. 10 etc.]

Exerc i se

1. I f "P2 = 9 2 4 0 , find n.

2. I f 1 0 p - 7 2 0 , find r.

when r

Work ( 0

CO

Soln:

Permutations & Combinations 643

Answers

n\ 1. Hint: Given, " A = 9240 * 7 = 9 2 4 0 j \n - 3 J!

or, ( - l X - 2 ) = 9240 = 2 2 x 2 1 x 2 0

.'. n = 22 [Here 9240 = 2 x 2 x 2 x 3 x 5 x 7 x 1 1 , greatest of those factors is 11 but it does not serve our purpose, there-fore try with 22]

10! 2 Hint:Given, 1 0 P r = 720 ' ( 1 0 _ , . ) = 7 2 0

.-. 10 x 9 x 8 x ...torfactors = 720= 10 x 9 x 8

.\3

Rule 2 Problems based on formation of numbers with digits

when repetition of digits is not allowed.

Working Rule (i) First of all decide of how many digits the required

numbers will be. (ii) Then fill up the places on which there are restrictions

and then apply the formula " p for filling up the

remaining places with remaining digits.

thousands place hundreds place tens place units place

Illustrative Example Ex.: How many numbers of four digits can be formed with

the digits 1,2,3,4, and 5? ( i f repetition of digits is not allowed).

Soln:

rlere n = number of digits = 5 and r = number of places to be filled up = 4

I . J . J . ^ . ^ L ^ - l VT;".',-;..'! .-. Required number = P 4 = = 5 x 4 x 3 x 2 = 120

Exercise 1. How many numbers between 400 and 1000 can be made

with the digits 2,3,4, 5,6 and 0? a) 60 b)70 c)40 d) 120

2. Find the number of numbers between 300 and 3000 that can be formed with the digits 0,1,2,3,4 and 5, no digits being repeated in any number. a) 90 b) 120 c)160 d) 180

3. Ho/ many even numbers of four digits can be formed with the digits 0, 1,2, 3, 4, 5 and 6; no digit being used more than once? a)300 b)140 c) 120 d)420

4. How many numbers of four digits greater than 23Q0 can

be formed with the digits 0, 1,2, 3, 4, 5 and 6; no digit being repeated in any number? a) 480 b)560 c)660 d)580

5. How many positive numbers can be formed by using any number of the digits 0, 1,2,3 and 4; no digit being repeated in any number? a) 360 b)260 c)620 d)280

Answers 1. a; Hint: [Here nothing has been given about repetition

of digits, therefore, we will assume that repetition of digits is not allowed.] Any number between 400 and 1000 must be of three digits only.

4 or 5 or 6

3 ways P2 ways

Since the number should be greater than 400, there-fore, hundreds place can be filled up by any one of the three digits 4, 5 and 6 in 3 ways. Remaining two places can be filled up by remaining

five digits in 5 P2 ways.

.-. Required number = 3 51

i p = 3 x - = 60 2 3!

2. d; Hint: Any number between 300 and 3000 must be of 3 or 4 digits. Case I : When number is of 3 digits.

3 or 4 or 5

3 ways 5 />2 ways Hundreds place can be filled up by any one of the three digits 3,4 and 5 in 3 ways. Remaining two places can be filled up by remaining

five digits in 5 P2 ways.

.. Number of numbers formed in this case = 3 x 5 p =

5! 3 x - = 6 0 . Case I I : When number is of 4 digits

1 or 2

1

2 ways P3 ways

Thousands place can be filled up by any one of the

644 P R A C T I C E B O O K O N Q U I C K E R M A T H S

two digits 1 and 2 in 2 ways and remaining three places can be filled up by remaining five digits in 5 P3 ways. .-. Number of numbers formed in this case

, 5! = 2 * 5 P 3 = 2 x = 120

.-. Required number = 60 + 120 = 180 3. d; H in t : Each even number must have 0 ,2 ,4 or 6 in its

units place. Here total number of digits = 7.

0 or 2 or 4 or 6

[When 0 occurs at units place there is no restriction on other places and when 2 or 4 or 6 occurs at units place there is restriction on thousands place as 0 can not be put at thousands place.] Case I : When 0 occurs at units place

0

i i 6 P 3 ways 1 way

Units place can be filled up by 0 in 1 way and remaining three places can be filled up by remaining 6 digits

in 6 P 3 ways. .-. Number of numbers formed in this case =

l x 6 P 3 = = 120 3 3!

Case I I : When 0 does not occur at units place.

any one of remaining six digits except zero 2 or 4 or 6

i 5 ways

i 5 P2 ways

I 3 ways

5.b;

Units place can be filled up by any one of the three digits 2,4 and 6 in 3 ways. Thousands place can be filled up by any one of the remaining six digits except zero in 5 ways. Remaining two places can be filled up by remaining

five digits in 5 P2 ways. .-. Number of numbers formed in this case

51 = 5 x3 x 5 p 2 =15 x - = 300

.-. Required number = 120 + 300 = 420 4. b; H in t : [Since number must be of four digits and greater

than 2300, therefore any one of the five digits 2 ,3 ,4 , 5 and 6 wil l occur at thousands place. When any one

of 3,4,5,6 occurs at thousands place the number will be definitely greater than 2300 but when 2 occurs at thousands place there wi l l be also restriction on hundreds place to make the number greater than 2300.] Case I : When 2 occurs at thousands place:

2 3 or 4 or 5 or 6

I I 1 1 way 4 ways 5 P 2 ways Thousands place can be filled up by 2 in 1 way and hundreds place can be filled up by any one of the four digits 3,4, 5 and 6 in 4 ways. Remaining two places can be filled up by remaining

five digits in 5 P 2 ways. .-. Number of numbers formed in this case

= ] x 4 x 5 p , = 4 x = 80 2 3!

Case I I : When anyone of 3,4,5 and 6 occurs at thousands place:

3 or 4 or 5 or 6

I I 4 ways 6 P 3 ways

Thousands place can be filled up by any one of the four digits 3,4,5 and 6 in 4 ways and remaining three

places can be filled up by remining six digits in 6 P3 ways. .-. Number of numbers formed in this case

. 4 x 6 A = 4 x = 480 3 3!

.-. Required number = 80 + 480 = 560 H i n t : Case I : When number is of five digits:

1 or 2 or 3 or 4

1 i 4 ways 4 P 4 ways

Ten thousands place can be filled up by any one of the four digits 1,2,3 and 4 in 4 ways and the remaining four places can be filled up by the remaining four digits in 4 P4 ways.

.-. Number of numbers formed in this case = 4 x 4 p 4

Case I I : When number is of four digits.

1 or 2 or 3 or 4

i i 4 ways 4 P 3 ways

646

be 1 or 2 or 3.

1 or 2 or 3 0 or 1 or 2 or 3 or 4

1 i i i 3 ways 5 ways 5 ways 5 ways

Thousands' place can be filled up by any one of the three digits 1,2 and 3 in 3 ways. Hundreds', tens' and units' places can each be filled by any one of the five digits 0, 1,2,3 and 4 in 5 ways. .-. Required number = 3 x 5 x 5 * 5 = 375

Rule 4 Theorem: If there are two groups A and B consisting of'm' and 'n' things respectively, then the number of ways in which

no two of group B occur together are given by Pn x ml).

Provided that n

Permutations & Combinations 647

pers.

These 9 papers can be arranged in 9 P9 = 9 ! ways

But these two papers can be arranged among them-selves in 2! ways. .-. number of arrangements when the best and the worst papers do not come together

= 101-9! x 2 ! = 9 ! ( 1 0 - 2 ) = 8 x9! . Quicker Method: Applying the above theorem, we have, the required number of ways = ( 1 0 - 2 ) x ( l 0 - l ) !

= 8 x 9 1 = 8 x 9 ! Note: The number of ways in which 'n ' books may be ar-

ranged on a shelf so that two particular books shall not be together is [(n - 2) x (n - 1)!]

Exercise 1. In how many ways can 12 examination papers be ar-

ranged so that the best and the worst papers never come together. a) 1 0 x 1 1 ! b) 1 2 x 1 1 ! c) 10x12! d) 1 0 ! x l l !

2. In how many ways can 15 examination papers be ar-ranged so that the best and the worst papers never come together. a) 13! x 14! b) 13x10! c) 13x14! d)Noneofthese

3. Find the number of ways in which 21 books may be arranged on a shelf so that the oldest and the newest books never come together. a) 19! x 20! b) 1 9 x 2 1 ! c) 19 x 20! d) Can't be determined

Answers l.a 2.c 3.c; Hint: See Note.

Rule 6 Theorem: There are 'm' boys and 'n'girls. The no. of ways in which they can be seated in a row so that all the girls do

not sit together are given by [(m + ) - (m +1) x ri\Note: This rule is different from the Rule-I. In Rule-I, "o

two occur together" is given whereas in this rule "all the girls do not sit together" is given.

Illustrative Example Ex.: There are 5 boys and 3 girls. In how many ways can

they be seated in a row so that all the three girls do not sit together.

Soln: Detail Method: Total number of persons = 5 + 3 = 8 When there is no restriction they can be seated in a row in 8! ways. But when all the three girls sit together, regarding the three girls as one persons, we have only 5 + 1 = 6 persons. These 6 persons can be arranged in a row in 6! ways.

But the three girls can be arranged among themselves in 3! ways .-. number of ways when three girls are together

= 6! x 3 ! .-. Required number of ways in which all the three

girls do not sit together = 8 ! - 6 ! x 3! = 6! (8 * 7 - 6 ) = 50x61=36000.

Quicker Method: Applying the above theorem, we have the required no. of ways = (5 + 3)! (5 + 1)! x 3!

= 8!-6! x 3! = 5 0 x 6 ! =36000. Note: There are'm' boys and 'n ' girls. The no. of ways in

which they can be seated in a row so that all the boys do not sit together are given by [(m + n)! - (n + 1)! x m!] ways.

Exerc i se 1. There are 3 boys and 2 girls. In how many ways can they

be seated in a IOW so that all the three boys do not sit together. a)72 b)42 c)172 d) 190

2. There are 8 boys and 4 girls. In how many ways can they be seated in a row so that all the girls do not sit together, a) 1320x9! b) 1296 x 9! c) 1344x9! d) 1296x 12!

3. There are 9 boys and 5 girls. Find the no. of ways in which they can be seated in a row so that all the boys do not sit together. a)240240x9! b)240240x5! c)240234x9! d)240236x9!

Answers l .a 2 .b 3. c; Hint: See Note.

Rule 7 Theorem: The number of ways in which m boys and'm' girls can be seated in a row so that boys and girls are alter-nate are given by 2(m! ml) ways.

Illustrative Example Ex.: In how many ways 4 boys and 4 girls can be seated in

a row so that boys and girls are alternate? Soln: Detail Method:

Case I : When a boy sits at the first place: Possible arrangement will be of the from

B G B G B G B G

Now there are four places namely 1 st, 3rd, 5th and 7th for four boys, therefore, four boys can be seated in 4! ways. Again there are four places namely 2nd, 4th, 6th and 8th for four girls. .-. four girls can be seated in 4! ways. .-. Number of ways in this case = 4! 4!

648 PRACTICE BOOK ON QUICKER MATHS

Case I I : When a girl sits at the first place, possible arrangement will be of the form

G B G B G B G B

.-. Number of arrangements in this case = 4! 4!

.-. Requirednumber = 4!4!+4!4!=2(4!4!)= 1152. Quicker Method: Applying the above theorem, we have

the requ ired answer = 2 (4! 4!) = 1152. Exerc ise 1. In how many ways 3 boys and 3 girls can be seated in a

row so that boys and girls are alternate? a) 9 b)36 c)72 d) Data inadequate

2. In how many ways 5 boys and 5 girls can be seated in a row so that boys and girls are alternate? a) 14400 b) 28800 c) 28000 d) None of these

3. In how many ways 2 boys and 2 girls can be seated in a row so that boys and girls are alternate? a) 4 b)2 c)8 d) 16

Answers l .c 2.b 3.c

Rule 8 Theorem: The number of ways in which m boys and (m -1) girls can be seated in a row so that they are alternate is given by [m! (m - 1)!] ways. Illustrative Example Ex.: In how many ways 4 boys and 3 girls can be seated in

a row so that they are alternate? Soln: Detail Method: Possible arrangement wil l be of the

form

B G B G B G B

There are four places namely 1 st, 3rd, 5th and 7th for four boys. .-. Four boys can be seated in 4! ways. Again there are three places namely 2nd, 4th and 6th for three girls. .-. Three girls can be seated in 3! ways .-. Requird number = 4! 3! = 144 Quicker Method: Applying the above theorem, we have the required answer = 4! 3! = 144.

Exerc ise 1. In how many ways 10 boys and 9 girls can be seated in

a row so that they are alternate? a)10!9! b) 10111! c )9 ! l l ! d) Data inadequate

2. In how many ways 8 boys and 7 girls can be seated in a row so that they are alternate? a)8!7! b)2(8!7!) c)8!8! d)8!9!

3. In how many ways 5 boys and 4 girls can be seated in a

row so that they are alternate? a)2(5!4!) b)4!4! c)2(4!4!) d)5!4!

Answers l .a 2.a 3.d

Rule 9 Theorem: ..< ..unber of ways in which m persons of a particular group, caste, country etc. andm persons of the other group, caste, community, country etc can be seated along a circle so that they are alternate, given by [m! (m -1)!J ways. Illustrative Example Ex.: In how many ways can 5 Indians and 5 Englishmen

be seated along a circle so that they are alternate? Soln: Detail Method: 5 Indians can be seated along a circle

in 4! ways [See Note in Rule - 10]. If the Englishmen sit at the places indicated by cross x then Indians and Englishmen will be alternate.

1

Now there are 5 places for 5 Englishmen. .-. 5 Englishmen can be seated in 5! ways. .-. Required number = 4! 5!. Quicker Method: Applying the above theorem, we have, the required number = 4! 5!.

Exerc ise 1. In how many ways can 4 Indians and 4 Englishmen be

seated along a circle so that they are alternate? a)2(4!3!) b)4!4! c)4!3! d)4!5!

2. In how many ways can 6 Indians and 6 Englishmen be seated along a circle so that they are alternate? a)6!5! b)6!6! c)2x6!5! d) None of these

3. In how many ways can 8 Indians and 8 Englishmen be seated along a circle so that they are alternate? a)8!8! b)8!9! c)8!7! d) None of these

Answers l .c 2.a 3.c

Rule 10 Theorem: A round table conference is to be held between n delegates. The no. of ways in which they can be seated so thatm particular delegates always sit together are given by f(n - m)! x ml] ways. Il lustrative Example Ex.: A round table conference is to be held between 20

660 PRACTICE BOOK ON QUICKER MATHS

a) 276 b)286 c)296 d) Can't be determined 2. Find the no. of triangles that can be formed with 14 points

in a plane of which no three points are collinear. a) 346 b)364 c)384 d)464

3. Find the no. of triangles that can be formed with 15 points in a plane of which no three points are collinear. a) 454 b)455 c)544 d)445

Answers l .b 2.b 3.b

Rule 20 Theorem: n students appear in an examination. The num-ber of ways the result of the examination can be announced

are given by (2)".

IIlustrativeExample Ex.: 6 students appear in an examination. In how many

ways can the result be announced? Soln: Detail Method: Each student can either pass or fail in

the examination. So, there exists 2 possibilities for each of the 6 students in the result. .-. total number of ways for the result" (2) 6 = 64 Quicker Method: Applying the above theorem, we have

the required answer = (2) 6 = 64 .

Exercise 1. 4 students appear in an examination. In how many ways

can the result be announced? a) 15 b) 16 c)17 d) None of these

2 5 students appear in an examination. In how many ways can the result be announced? a) 32 b)33 c) 31 d) Data inadequate

3. 7 students appear in an examination. In how many ways can the result be announced? a) 126 b)127 c)129 d) 128

Answers l .b 2.a 3.d

Rule 21 Theorem: n' matches are to be played in a chess tourna-ment. The number of ways in which their results can be

decided are gien by (3)" ways.

Illustrative Example Ex: 3 matches are to be played in a chess tournament. In

how many ways can their results be decided? Soln: The result of each of the 3 matches can be in three

ways namely win, draw or loss. total no. of ways in which results of 3 matches can

be decided = (3^ = 27.

Exercise 1. 2 matches are to be played in a chess tournament. In

how many ways can their results be decided? a) 27 b)9 c)8 d) None of these

2. 4 matches are to be played in a chess tournament. In how many ways can their results be decided? a) 81 b) 16 c)27 d)64

3. 5 matches are to be played in a chess tournament. In how many ways can their results be decided? a) 343 b)243 c) 128 d) None of these

Answers l . b 2.a 3.b

Rule 22 Theorem: A badminton tournament consists of 'n' matches.

(i) The number of ways in which their results can be

forecast are given by ( 2 ) " ways.

(ii) Total number of forecasts containing all correct re-sults or all wrong results are given by 1.

Illustrative Example Ex: A badminton tournament consists of 3 matches.

(i) In how many ways can their results be forecast? (ii) How many different forecasts can contain all wrong results. (iii) How many different forecasts can contain all cor-rect results?

Soln: Each badminton match can be decided in only 2 ways ie win or loss for a particular team. .-. Total no. of ways the results of 3 matches can be

forecast = 2 3 = 8 Result of each match can be forecast wrong in only 1 way. .-. Total no. of forecasts containing all wrong results

Similarly, result of each match can be forecast correct in only 1 way.

.-. Total no. of forecasts containing all correct results

= (1) 3 =1

Exercise 1. A badminton tournament consists of 4 matches. In how

many ways can there results be forecast? a) 16 b)32 c)15 d) 31

2. A badminton tournament consists of 5 matches. In how many ways can their results be forecast? a)32 b)31 c)33 d)64

3. A badminton tournament consists of 7 matches. In how many ways can their results be forecast? a)64 b)70 c) 128 d) 127

Permutations 85 Combinations 6 6 1

Answers l .a 2.a 3.c

Rule 23 Theorem: The Indian hockey team is to play 'n' matches for the world cup.

(i) The number of differentforecasts that will contain all

correct results is ( l ) " or 1.

(ii) The number of different forecasts that will contain all wrong results are (2)".

Illustrative Example Ex.: The Indian hockey team is to play 4 matches for the

world cup. (i) How many different forecasts will contain all cor-rect results? (ii) How many different forecasts will contain all wrong results?

Soln: Result of each hockey match can be decided in 3 ways ie win, loss or draw. (i) Only in 1 way we can correctly predict each match

ie 'forecast' and the actual result are the same. Hence, total no. of ways to predict all 4 matches

correctly = ( l ) 4 = 1. (ii) For each match the prediction can go wrong in 2

ways. For example, prediction for any match is win but the actual result is either loss or draw. .-. total no. of forecasts containing all wrong

results = (2) 4 =16.

Exercise 1. The Indian hockey team is to play 3 matches for the

world cup. How many different forecasts will contain all correct results? a)8 b) 16 c ) l d)32

2. The Indian hockey team is to play 8 matches for the world cup. (i) How many different forecasts will contain all correct

results? a) 8 b ) l c)256 d) None of these

(ii) How many different forecasts will contain all wrong results? a) 128 b)256 c)512 d) None of these

Answers l .c 2.(i)b;(ii)b

Miscellaneous 1. From 4 officers and 8 jawans in how many ways can be 6

chosen to include at least one officer. a) 896 b)986 c)886 d)996

2 From a group of 6 men and 4 women a committee of 4 persons is to be formed.

(SBI Bank PO Exam 2001) (i) In how many different ways can it be done so that the

committee has at least one woman? a)210 b)225 c)195 d) 185

(ii) In how many different ways can it be done so that the committee has at least 2 men? a)210 b)225 c) 195 d) 185

Answers l .a; Hint: Detail Method:

No. of officers No. of jawans No. of ways

4 Case I 1

Case I I 2

Caselll 3

Case IV 4

5

4

3

Cj x 6 C 5 = 224

' C 2 x 8 C 4 =420

' C 3 x 8 C 3 =224

4 C 4 x 8 C 2 =28

.-. Required number = 224 + 420 + 224 + 28 = 896

Quicker Method: Applying the above theorem, we have,

x = 4,y = 8andn = 6

The value of

XCX x yCn_x = C 4 x C 6 _ 4 = C 4 x C 2

Now, from the above theorem,

Required answer

= 4 C , x 8 C 5 + 4 C 2 x 8 C 4 + 4 C 3 x 8 C 3 + 4 C 4 x 8 C 2

= 224 + 420 + 224 + 28 = 896

2. (i) c; Hint: Required no. of ways

= 4 C , x 6 C 3 + ' 4 C 2 x 6 C 2 + 4 C 3 x 6 C, + 4 C 4

6 x 5 x 4 4x3 6x5 4x3x2 4x x6 + l

1x2x3 1x2 1x2 1x2x3

= 80 + 90 + 24 + 1 = 195

(ii) d; Hint: Required no. of ways

= 6 C 2 x 4 C 2 + 6 C 3 x 4 C j + 6 C 4

6x5 4x3 6 x 5 x 4 6x5x4x3 -x = x4 + -1x2 1x2 1x2x3 1x2x3x4

= 90 + 80+ 15=185