SOLVING WORD PROBLEMS INVOLVING PERMUTATION

Problem 1 :

There are six greyhounds in a race :

Spot, Fido, Bowser, Mack, Tuffy, William.

How many different 1^st, 2^nd, 3^rd place orders of finish are possible?

Solution :

Total number of greyhounds participated in the race = 6

From 6 greyhounds only 3 of them will get 1^st place, 2^nd place and 3^rd place.

= ⁶P₃

= 6!/(6-3)!

= 6!/3!

= (6 x 5 x 4 x 3!)/3!

= 120 ways

Problem 2 :

If a chess association has 16 teams, in how many different ways could the top 8 positions be filled on the competition ladder ?

Solution :

Number of teams participated = 16

Number of teams to be selected to fill top 8 positions.

= ¹⁶P₈

= 16!/(16-8)!

= (16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8!)/8!

= (16 x 15 x 14 x 13 x 12 x 11 x 10 x 9) ways

Problem 3 :

Suppose you have the alphabet blocks A, B, C, D and E and they are placed in a row.

a) How many ways could you have ?

b) How many ways we have to end in C.

c)  How many ways to have a form of

___ A ___ B ___

d) How many begin and end with a vowel, A or E ?

Solution :

Number of letters to be arranged = 5

a)  All letters to be chosen and have to arrange them.

= ⁵P₅

= 5! / (5-5)!

= 5!/0!

= 5!/1

= 120 ways

b)  In the last place, we fix C. Excluding 1 letter, we can choose 4 letters and arrange.

= 4! x 1

= 24 ways

d)  ___ A ___ B ___

To fill the 1^st space, we have 3 ways.

To fill the 2^nd space, we have 2 ways.

and

To fill the 3^rd space, we have 1 way.

= 3 x 1 x 2 x 1 x 1

= 6 ways

Problem 4 :

In how many ways can 5 different books be arranged on a shelf ?

Solution :

Number of books available = 5

Number of books to be chosen and arranged = 5

= ⁵P₅

= 5! / (5-5)!

= 5!/0!

= 5!/1

= 120 ways

Problem 5 :

How many 3 digit numbers can be constructed from the digits 1, 2, 3, 4, 5, 6, 7 each digit may be used :

a) As often as desired

b)  Only once

c)  Only once and the number is odd. 

Solution :

Total number of digits = 7

a) To create a 3 digit number, all 7 digits are having equal chances to be filled.

Since there is no restriction, we have

= 7 x 7 x 7

= 343 ways

b)  Only once

So, number of ways = ⁷P₃

= 7!/(7 - 3)!

= 7!/4!

= (7 x 6 x 5 x 4!)/4!

= 210 ways

c)  Only once and the number is odd.

Since the number to be created is odd, the last place should be filled with 1, 3, 5 and 7.

___ x ___ x 4 options

Excluding 1 value, we have already filled, we have 6 options in the 10's place.

___ x 6 x 4 options

Excluding 2 values that we have already filled, we have 5 options in the 1's place.

= 5 x 6 x 4

= 120

So, total ways is 120.

Problem 6 :

3 digit number are constructed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 using each digit at most once. How many such numbers :

a) Can be constructed 

b)  end with 5

c) end with 0.

d) are divisible by 5.

Solution :

a) The required 3 digit number may be anything but that will not start with 0. So, excluding 0, we have 9 options are available to fill 100's digit.

In 10's place, we can put any numerical value including 0. So, we have 9 options again.

In 1's place, we have 8 options.

So, the required number of ways are 

= 9 x 9 x 8

= 648

b)  Should end with 5 and does not start with 0,

= 8 x 8 x 1

= 64

c) Should end with 0, so we have 1 option for 100's place. Then, we have to fill 1's place.

= 9 x 8 x 1

= 72

d) The number divisible by 5, when it ends with 0 or 5.

If the number ends with 0, we have 

= 9 x 8 x 1

= 72 ----(1)

If the number ends with 5, we have 

= 8 x 8 x 1

= 64 ----(2)

(1) + (2)

= 72 + 64

= 136 ways

Problem 7 :

In how many ways can 5 different books be arranged on a shelf if 

a) there is no restrictions

b)  books X and Y must be together

c)  Books X and Y never together.

Solution :

a) 5 books should be arranged, since there is no restrictions

= 5 x 4 x 3 x 2 x 1

= 120 ways

b)  X and Y both can be considered as one group. So, we can consider those two books as one unit. Then we have 4 units 

= 4!

2 books X and Y can be shuffled in 2! ways.

= 4! x 2!

= 48 ways.

c)  Books X and Y never together.

The two books are never together, so the total number of ways can be found out by subtracting the number of cases when 2 books are always together from the total number of ways of arranging the 5 different books.

= 120 - 48

= 72 ways