Paper 4: Fundamentals of Business Mathematics & Statistic

2.42 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Algebra

or,

( )( )( )( )

( )

( )( )

( )

n n 1 n 2 n 3 n 4! 30 n n 1 n 2!

n 4! n 2!

− − − − = × − −

− −

or, n(n – 1) (n – 2) (n – 3) = 30 × n(n – 1) or, (n – 2) (n – 3) = 30

or, n^2 – 5n – 24 = 0 or, (n – 8) (n + 3) = 0

or, n = 8, – 3 (inadmissible)

Example 66 : Solve for n given

n 5

n 3

P 2

P = 1

Solution :

n 5

n 3

P 2

P =^1 or, nP 2 P^5 = ×n^3

or, ( ) ( )

n! = 2× n!

n- 5! n- 3! or, ( ) ( )

1 2^1

= ×n 3 n 4− −

or, n^2 – 7n + 10 = 0 or, n = 5, 2 (inadmissible).

Example 67: In how many ways 6 books out of 10 different books can be arranged in a book-self so that

3 particular books are always together?

Solution :

At first 3 particular books are kept outside. Now remaining 3 books out of remaining 7 books can be

arranged in^7 P 3 ways. In between these three books there are 2 places and at the two ends there are 2

places i.e. total 4 places where 3 particular books can be placed in^4 P 1 ways. Again 3 particular books can

also be arranged among themselves in 3! ways.

Hence, required no. of ways =^7 P P 3! 3 ×^41 × =7! 4!4! 3!× ×3! = 7.6.5.4.3.2.1 = 5040.

Example 68: In how many ways can be letters of the word TABLE be arranged so that the vowels are

always (i) together (ii) separated?

Solution :

(i) In the word there are 2 vowels, 3 consonants all different. Taking the 2 vowels (A, E) as one letter we

are to arrange 4 letters (i.e. 3 consonants + 1) which can be done in 4! ways. Again 2 vowels can be

arranged among themselves in 2! ways.

Hence, required number of ways = 4! × 2! = 48.

(ii) Without any restriction (i.e. whether the vowels, consonants are together or not) all the different 5

letters can be arranged in 5! ways. Arrangement of vowels together is 48 (shown above)

Hence, Required number of ways = 5! – 48 = 120 – 48 = 72.

Example 69 : Find the how many ways can be letters of the PURPOSE be rearranged–

(i) keeping the positions of the vowels fixed ;

(ii) without changing the relative order to the vowels and consonants.

Solution :

(i) In the word, there are 3 vowels and 4 consonants. Since the positions of all vowels fixed, we are to

rearrange only 4 consonants, in which there 2 P, so the arrangement is