Paper 4: Fundamentals of Business Mathematics & Statistic

2.40 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Algebra
nPn = n (n – 1) (n – 2) ...... to n factors
= n (n – 1) (n – 2) ..... 3 × 2 × 1 [putting r =n]
Again, nPn–1= n (n – 1) (n – 2) .... 3 × 2 [putting r = n – 1]
∴ nPn = nPn – 1
Factorial notation :
The continued product of first n natural numbers, i.e., 1, 2, 3, .... (n – 1) n, is generally denoted by the symbol
nor n! which is read as ‘factorial n”.
Thus,5! = 1× 2 × 3 × 4 × 5 =120 ( = 5 × 4!) = (5 × 4 × 3!)
7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = (1 × 2 × 3 × 4 × 5 × 6 ) × 7 = 7 × 6!
n! = 1 × 2 × 3 ×.....× (n – 2) × (n – 1) n = nPn ; O! = 1
Obs. nPr = n (n – 1) (n – 2) ... (n – r + 1)
( )( ) ( )( )
( ) ( )

n n 1 n 2 ..... n r 1 n r! n!

n r! n r!

= − − − + − =

− −

Permutation of things when they are not all different :

To find the number of permutation of n things taken all together, the things are not all different.

Suppose that n things be represented by n letters and p of them be equal to a, q of them be equal to b and

r of them be equal to c and the rest be all different. Let X be the required number of permutations.

( )

( ) ( ) ( )

X n!

∴ = p! q! r!

Cor. The above method is also applicable when more than three letters are repeated.

Example 58 : Word PURPOSE can be arranged all together in

7!

2! as 2P’s are there.

Permutations of things which may be repeated :

If n different things are taken r at a time, in which any item can be repeated without any restriction, the

total number of possible arrangements is nr.

Permutations in a ring or in a circle :

When things are arranged in a row, we find two ends in each arrangement, while when the things are

arranged in circle, there is no such end.

Thus the number of ways in which n different things can be arranged in a circle taking all together is

(n – 1) !, since any one of the things placed first is fixed and remaining (n – 1) things can now be arranged

in (n –1)! ways.

Example 59 : 21 boys can form a ring in (21 – 1)! = 20! ways, If, again the distinction between the clockwise

and counter-clockwise arrangements is not made, then the number of ways is ½ (n – 1) !.

Example 60: 10 different beads can be placed in a necklace in^12 ×(10 1 !−) =^12 ×9! ways.

Restricted Permutation :

(i) The number of permutations of n different things taken r at a time in which p particular things never

occur is n –pPr.