SOLVING WORD PROBLEMS ON FACTORIALS

Problem 1 :

In how many ways is it possible for 15 students to arrange themselves among 15 seats in the front row of an auditorium?

Solution :

Number of persons available to sit in the first chair = 15

Number of persons available to sit in the second chair = 14

It keeps going in this way,

So, total number of ways to sit 15 students = 15!.

Problem 2 :

There are 8 greyhounds in a race. How many different orders of finish (first place through eighth place) are possible?

Solution :

Number of greyhounds will take the 1^st position = 8

Number of greyhounds will take the 2^nd position = 7

Continuing in this way, we get

= 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

= 40320

Problem 3 :

There are ten candidates for a job. The search committee will choose four of them, and rank the chosen four from strongest to weakest. How many different outcomes are possible?

Solution :

Total number of candidates = 10

To choose strongest candidate, there are 10 options. To choose the second strongest candidate, there are 9 options available. Continuing in this way, we get

= 10 x 9 x 8 x 7

= 5040

So, total number of ways to select 4 candidates, we have 5040 ways.

Problem 4 :

There are 8 horses in a race. If all we are concerned with are the first, second and third place finishers (the trifecta), how many different orders of finish are possible?

Solution :

Total number of horses = 8

Number of persons who are having chances to reach the first position = 8

Number of persons who are having chances to reach the second position = 7

Number of persons who are having chances to reach the third position = 6

= 8 x 7 x 6

= 336

So, the total number of ways is 336.

Problem 5 :

Suppose we are going to use the symbols {a, b, c, d, e, f, g, h} to form a 5-character "password" having no repeated characters. How many different passwords are possible?

Solution :

Given set = {a, b, c, d, e, f, g, h}

Total number of distinct character = 8

The number of options to fix the 1^st letter of password = 8

The number of options to fix the 2^nd letter of password = 7

The number of options to fix the 3^rd letter of password = 6

The number of options to fix the 4^th letter of password = 5

The number of options to fix the 3^rd letter of password = 4

= 8 x 7 x 6 x 5 x 4

= 6720

So, total number of ways = 6720.

Problem 6 : 

There are six greyhounds in a race:

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

We are concerned about who finishes first, second and third. How many different 1^st, 2^nd, 3^rd orders of finish are possible?

A) 120    B)  216      C) 18       D) 15

Solution :

Number of greyhounds = 6

Number of greyhounds to fill the 1^st position = 6

Number of greyhounds to fill the 2^nd position = 5

Number of greyhounds to fill the 3^rd position = 4

= 6 x 5 x 4

= 120

Problem 7 :

Homer, Gomer, Plato, Euclid, Socrates, Aristotle, Homerina and Gomerina form the board of directors of the Lawyer and Poodle Admirers Club. They will choose from amongst themselves a Chairperson, Secretary, and Treasurer.

No person will hold more than one position. How many different outcomes are possible?

A) 336    B) 24     C) 512     D) 21

Solution :

Number of persons = 8

Number of persons who occupies the position

of Chairperson = 8

Number of persons who occupies the position

of Secretary = 7

Number of persons who occupies the position

of Treasurer = 6

= 8 x 7 x 6

= 336

Problem 8 :

A flutter on the horses There are 7 horses in a race

a)  In how many different orders can the horses finish? 

b) How many trifectas (1^st, 2^nd and 3^rd) are possible?

Solution :

a)  Number of available options = 7 x 6 x 5 x 4 x 3 x 2 x 1

= 5040

b) Number of ways to choose 1st, 2nd and 3rd position :

= 7 x 6 x 5

= 210

Problem 9 :

In how many ways can 5 boys and 4 girls be arranged on a bench if

a) there are no restrictions?

b) boys and girls alternate?

Solution :

a)  Without restriction :

Number of boys = 5, number of girls = 4

Total number of members = 5 + 4

= 9

Without restrictions, we have = 9! ways.

b)  Boys and girls to be seated in alternate positions.

= 5! x 4! ways

Problem 10 :

How many arrangements of the letters of the word REMAND are possible if:

a) there are no restrictions?

b) they begin with RE?

c) they do not begin with RE?

Solution :

Given word : REMAND

Number of letters in the word = 6

a)  Without restrictions :

Number of ways, that letters can be arranged = 6!

b) they begin with RE?

R E ___  ___   ___  ___

Number of words to be arranged = 4

Number of ways available = 4!

c) they do not begin with RE?

Number of words can be created, that doesn't start with RE

= 6! - 4!

= 4!(6 x 5 - 1)

= 4! x 29

Problem 11 :

How many arrangements are there of the letters of the word SCROOGE?

In how many of these are the O' s together?

Solution :

Total number of letters in the given word = 7

To arrange those 7 letters, we have 7! ways. Here 2 terms cannot be separated.

Total number of ways = 7!/2!

= (7 x 6 x 5 x 4 x 3 x  2 x 1)/(2 x 1)

= 2520

Problem 12 :

How many arrangements are there of the letters from ABACUS in which the A' s are together?

Solution :

Total number of letters in the given word = 6

Excluding two A's, we have B, C, U and S. We consider  4 + 1 = 5 units.

Total number of ways = 5!

= 5 x 4 x 3 x 2 x 1

= 120