Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 2.41

Keeping aside the p particular things, fill up the r places with the remaining n – p things.
Hence, number of ways = n–pPr.
Example 61 : In how many of the permutations of 8 things taken 3 at a time, will two particular things never
occur?
Solution :
Here, n = 8, r = 3, p = 2,
Hence, Number of ways = n–pPr = 8–2P 3 =^6 P 3 = 120.
(ii) The number of permutations of n different things taken r at a time in which p particular things are
always present is n p−P P .r p− ×rp

Example 62 : In how many of the permutations of 8 things taken 3 at a time, will two particular things
always occur?
Solution :
Here, n = 8, r = 3, p = 2,

Number of ways = n p−P Pr p− ×rp=8 2− P P3 2− ×^32 =^6 P P 1 ×^32 = 6 × 3 = 18.

SOLVED EXAMPLES

Example 63 : Find the values of– (i)^7 P 5 (ii)^7 P 1 (iii)^7 P 0 (iv)^7 P 7
Solution :

(i)^75 ( )

P 7! 7! 7.6.5.4.3.2.1

= 7 5 !− =2 != 2.1 = 7.6.5.4.3 = 2520

(ii)^71 ( )

P 7! 7! 7 6! 7

= 7 1 !− =6 != ×6 !=

(iii)^70 ( )

P 7! 7! 1

= 7 0 !− =7 !=

(iv) ( )

(^7) P 7 7! 7! 7!
= 7 7 !− =0! 1!= = 7.6.5.4.3.2.1 = 5040
Example 64 : If nP 110, 2 = n.
Solution :

( )

( )( )

( )

nP 2 n! 110 or,n n 1 n 2! 110

n 2! n 2!

= = − − =

− −

or, n(n – 1) = 110 = 11 × 10 = 11 × (11 – 1) ∴ n = 11

Example 65 : Solve for n given nP 30 P 4 = ×n 2
Solution :

nP 4 = 30 × nP 2 or,
( ) ( )

n! 30 n!

n 4 !− = × n 2 !−