Theory and Questions Related to Permutation & Combination

PERMUTATIONS & COMBINATIONS

PERMUTATIONS AND COMBINATIONS

Multiplication principle:

A procedure is completed in two steps, A and B. B follows A. If A can be performed in m ways and B in n ways, then there are mn ways to complete the procedure.

Addition principle:

A procedure is such that it can be performed in A or in B way.

The procedure can be completed in m + n ways.

Example 1: Number of 4 digit numbers using 1, 2, 3, 4, 5:

(i) any digit can occur any number of times

ways:

(ii) repetition of digits is not allowed

ways:

Example 2: Number of 4 digit numbers using 0, 1, 2, 3, 4:


ways:


ways:

Page 1

S P Gupta. A - 12 (Opp. Shri Ram Mandir), Mathur Vaish Nagar, Tonk Road, Jaipur – 302029 Ph.: 2549422, 2554655, 9829055465.

IVIIIIII

5555

6255555


Example 3: Number of 4 digit even numbers using 1, 2, 3, 4, 5:

(i) any digit may be repeated any number of times

ways:


ways:

Example 4: Number of 4 digit even numbers using 0, 1, 2, 3, 4:


ways:


ways:or

ways:

Let a, b, c be three dissimilar objects.

1. Number of permutations of n dissimilar things = n !

2. Number of permutations of n dissimilar things taken r at a time =

(i).

(ii).

Page 2


IIVIIIII

2555

2505552 4or2

IIVIIIII

2234

482342 4or2

IIVIIIII

1234

242341 0


Given n dissimilar objects:Number of ways in which r objects can be selected

chosengroupedtaken

is denoted by the symbol

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Example 1: If 15 Pr = 2730, then find r.

Solution: 2730 = 15 x 14 x 13r = 3

Example 2: If 18 Cr = 18 C r + 2, find r P3.

Solution: 18 Cr = 18 Cr+2

Example 3: Words are formed using the letters of the word ‘JODHPUR’.

1. Total number of words = 7!

2. Total number of words in which P,U,R, are together = 5! 3! = 720(considering P, U, R to be one unit, we can arrange the 5 letters in 5! ways. P, U, R can change their positions within themselves in 3! ways).

3. Number of words in which P, U, R are not together = 7! – 5! 3! = 4320

(Total number of words – number of words in which P, U, R are together).

4. Number of words in which no two of P, U, R are together= 4! 5C3. 3! x x x x xPositions for J,O,D,Hx : Positions where P,U and R can be placedThe letters J,O,D,H can be arranged in 4! ways.We have 5 positions for placing 3 letters P, U, R. This can be done in 5C3. 3! ways.

Page 3



5. Number of words in which all the vowels are together and all the consonants are together = 2! (2! 5!)(Vowels within themselves can be arranged in 2! ways and consonants in 5! ways. The two groups of vowels and consonants can change their positions in 2! ways).

6. Number of words starting with J and ending with R = 5!

J R

(Positions of J and R are fixed. Therefore, we need to permute O, D, H, P and U only).

7. Number of words in which relative positions of vowels and consonants remain unchanged = 2! 5!

8. Number of words in which vowels occupy odd places = 4C2 2! 5!(There are four odd places: 1st, 3rd, 5th, and 7th. We have two vowels O and U. So we, first of all, select two positions and arrange vowels there. This can be done in 4C2 2! ways. Then, the remaining five places can be filled in 5! ways.)

9. Total number of 4 letter words when any letter can be repeated any number of times is equal to

ways:

10. Four letter words, using the letters of the word ‘JODHPUR’, are formed. The number of words which have at least one letter repeated is equal to .

11. Rank of the word ‘JODHPUR’ in the dictionary of the words formed with the letters of the word ‘JODHPUR’:Words starting with D or H:

ways:

Words starting with JD or JH:

ways:

Words starting with JODHPR:

J O D H P R U

Required rank =

Page 4


7777

1440!621234562

HorD

240!521234521

HorDJ


Let there are n objectsn1 objects are alike of one typen2 objects are alike of second type

.

.

.nk objects are alike of kth type

Number of permutaions of these n objects

Let there are n objects (all dissimilar). r objects are to be chosen with/without restrictions.

1. Without any restrictionNumber of ways = nCr

2. n1 particular objects are included in the selection

Number of ways =

3. n2 particular objects are not included in the selectionNumber of ways =

4. n1 particular objects are taken and n2 particular objects are not taken

Number of ways =

Example 4: Words are formed using the letters of the word ‘EXAMINATION’. Find (i). the number of words (ii). the number of four letter words

In the word ‘EXAMINATION’, we have

E – 1, X – 1, M – 1, T – 1, O – 1

A – 2, I – 2, N – 2

(i). Total number of words =

(ii). Four letter words:

Example 5: Words are formed using the letters of the word ‘JODHPUR’. Find the number of words in which P comes before U and U comes before R.

Let us designate each one of P, U and R by a single letter (say X). Now, we arrange X, X, X, J, O, D, H to form words.

Page 5


Cases Number of words

1. All the letters are distinct

2. Two letters are alike and two are different

3. Two letters are alike of one kind and two are of another kind

Total 2454


Total number of words =

In these words, if the first X (seen from the left) is replaced by P, the second X by U and the third X by R, then we get the words in which P comes before U and U before R. Hence the required number of words is 840.

Example 6: How many numbers greater than 10 lacs can be formed using 2,3,0,3,4,2,3.

Solution: All the given digits are to be used.Required number of numbers = (Total numbers formed including the numbers containing 0 at the left most position) – (total number of number of numbers starting with zero)

=

=

Example 7: There are 3 children, 4 women and 5 men. A group of 4 persons is to be formed containing atleast one women. Find the number of ways in which this can be done.

Solution: Total number of ways in which we can select 4 persons (without any restriction) =

Number of groups not containing any woman =

Required number of ways of selecting 4 persons =

1. Let there be n dissimilar objects.Selection of one of more objects can be done in 2n – 1 ways.

Proof: I. p objects may be selected in ways.

Total number of ways =

II. Since, one or more objects are to be taken, hence each object has two possibilities: (i) may be taken (ii) may not be taken.

Hence, the number of ways in which we may select or may not select the objects is equal to 2 x 2 x 2 x . . . n times i.e. 2n.This includes one way in which no object is taken.

Number of ways of selecting one or more objects is equal to 2n – 1.

2. Let there be n objects (not all are different).n1 objects are alike of first type.n2 objects are alike of second type. . . .nk

objects are alike of kth type..

Number of ways in which the objects may be selected is .

Example 8: 5 balls of different colors are given. In how many ways the balls may be selected?

Solution: Required number of ways = 25 –1= 31

Page 6



Example 9: 3 red, 4 white and 5 blue balls are given.(i). Number of ways in which balls may be chosen

= 119.(ii). Number of ways of selecting the balls when atleast one red ball is included in the selection = 3 ( 4 + 1 ) ( 5 + 1 )

= 90.(iii). Number of ways of selecting the balls when atleast one ball of each colour is included = 3.4.5 = 60.

Example 10: 3 red, 4 yellow, 5 blue balls are given. All the red balls are of different shades.(i). Number of ways of selecting one or more balls

= 239.(ii). Number of ways of selecting the balls when atleast one red ball is included = 210.

Example 11: There are four questions in a question paper. Each question has an alternative. In how many ways the question paper may be attempted?

Solution: Each question has three ways associated with it. 2 ways to do it and 1 way for not doing. Hence, the question paper may be attempted in 3 x 3 x 3 x 3 – 1 i.e. 80 ways.

Let there b n objectsn1 objects are alike of first type.n2 objects are alike of second type.

nk objects are alike of kth type.(n1 + n2 + . . . + nk = n).

Number of ways in which exactly r objects may be taken= Coefficient of xr in (x0 + x1 + . . . + ) (x0 + x1 + . . . + )

. . . (x0 + x1 + . . . + )

If there are at least r objects of each type i.e. , then

Number of ways in which exactly r objects may be taken.

= coeff.

Number of ways of dividing objects into groups:

n object (all dissimilar) are to be divided into k groups such that these groups contain n1, n2, n3, . . ., nk, objects (n1+ n2+ . . . + nk = n)

(i) If ni nj, i j (i.e. no two groups contain equal number of objects), then number of ways in which this can be done is

Page 7



(ii) If there be p groups (among these k groups) containing equal number of objects, then the number of ways is

Example 12: 12 dissimilar objects are to be put into groups.

Number of groupsNumber of objects in the groups

Number of ways of dividing objects into groups

3 3, 4, 5

2 6, 6

3 4, 4, 4

4 3, 3, 3, 3

4 2, 2, 2, 6

5 2, 2, 2, 3, 3

6 4, 4, 1, 1, 1, 1

Page 8



Example 13: Find number of ways in which 12 apples can be equally divided among 3 persons.

Solution: Apples are being considered to be dissimilar. 4 apples are to be given to each person. We may solve this problem in two different ways.

First approach: Take 4 apples and give it to one person. 8 apples are left now. Take 4 apples and give it to second person. Remaining 4 apples are given to the third person.

Required number of ways

Second approach: Divide 12 apples into 3 groups each containing 4 apples, and then permute the groups against the persons.

Required number of ways

Example 14: Five balls are to be placed in three boxes. Each box is capable of holding all the five balls. The balls are placed in the boxes in such a way that no box remains empty. Find the number of ways of placing the balls in the boxes.

Solution: We divide 5 balls into 3 groups in such a way that each group contains atleast one ball. Now, these groups are permuted against the boxes. Thus, number of ways in which this can be done is the number of ways of placing atleast one balls in the boxes.

(i). All the boxes are dissimilar and all the balls are dissimilar.

Number of balls in the groups

Number of ways of dividing balls into groups

Number of ways of placing the balls in the boxes

1, 1, 3 10 . 3! = 60

2, 2, 1 15 . 3! = 90

__________________________________________________________________

(ii) Balls are identical and boxes are dissimilar.




1, 1, 3 1

2, 2, 1 1

Balls are dissimilar and boxes are identical

Page 9






1, 1, 3

2, 2, 1

Balls are identical and boxes are also identical




1, 1, 3 12, 2, 1 1

__________________________________________________________________

Circular Permutations:

These arrangements are identical with respect to the objects but are distinct relative to the direction.

(i) No. of circular permutations of n dissimilar object = (n – 1)!(ii) No. of circular permutations of n dissimilar objects taken r at a time

Examples: (i). Number of ways in which n distinct beads can be stung into a necklace is

(ii). Number of ways in which 5 distinct objects can be given to 2 persons is 2 x 2 x 2 x 2 x 2

(iii). Number of ways in which 5 distinct objects can be distributed between 2 persons is 25 – 2.

(iv). Number of ways is which 5 identical objects can be given to 2 personsis equal to Coefficient of x5 in (1 + x + x2 + x3 + x4 + x5)2

(v). Number of ways in which 5 identical objects can be distributed between2 persons is equal to coefficient of x5 in (x + x2 + x3 + x4 + x5 )2

Number of rectangles on a chessboard Number of squares on a chessboard:

Page 10


a

a

c b

c

b a


Dimensions Number of squares

. .

. .

. .

Number of non – congruent rectangles on a chessboard:

Dimensions Number of rectangles 87

. . .

. . .

. . .1

Number of zeros at the end of 1000!

=

= 200 + 40 + 8 + 1 = 249

7 | 1000 C500 (True/False)

Hint: 1000 C500 =

Let

and are given by

Number of times we write 3 while writing numbers 1 to 1000:3 occurs exactly once in 3C1.9.9 numbers.3 occurs exactly twice in 3C2.9 numbers.3 occurs exactly thrice in 3C3 numbers.

So, the number of times, we write 3 while writing the numbers 1 to 1000 is 243 + + i.e. 300.

Page 11



B n parallel roads Find the number of shortest paths from A to B.

As long as a person moves eastward or southward he is on the shortest path. For going from A to B by a shortest path, one has to move n – 1 nodes eastward and m – 1 nodes southward. Let us denote eastward movement by one node by E and southward movement by one node by S.

Required number of shortest routes = Number of arrangements of n – 1 E’s and m – 1 S’s

Number of derangements of n objects

If n distinct objects are arranged in a row, then number of ways of selecting r consecutive objects is n – r + 1.

n members of a new club meet each day for a lunch at a round table. They decide to sit such that every member has different neighbours at each lunch. The number of days can this arrangement will last is equal to

11. n straight lines are drawn in the plane such that no two lines are parallel and no three lines are concurrent. The number of parts into which these lines divide the plane, is

12. Sum of numbers formed by using the digits 1,2,3,4,5 (repetition of digits is not allowed) : 1 occurs at the units place in 4! numbers. 1 occurs at the ten’s place in 4! numbers. 1 occurs at the hundred’s place in 4! numbers. 1 occurs at the thousand’s place in 4! numbers. 1 occurs at the ten thousand’s place in 4! numbers.

We have similar results for the other digits.So the required sum

Page 12


A

...

roadsparellelm

N

E


+

13. Number of factors (divisors) of 1800:

1800 = 23.32.52

A factor (divisor) of 1800 is of the form where

Total number of factors (divisors) of 1800 is (Number of proper divisors of 1800 is 36-2 i.e. 34)

Sum of the divisors of 1800 is equal to .

Page 13


Documents

Theory and Questions Related to Permutation & Combination