Paper 4: Fundamentals of Business Mathematics & Statistic

2.46 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Algebra

Total number of combinations :

To find the total number of combination of n different things taken 1, 2, 3 .... n at a time.

= nC 1 + nC 2 + nC 3 + ...... + nCn

Note. nC 1 + nC 2 + nC 3 + ..... + nCn = 2n –1

GROUPING :

(A) If, it is required to form two groups out of (m + n) things, (m ≠ n) so that one group consists of m things

and the other of n things. Now formation of one group represents the formation of the other group

automatically. Hence the number of ways m things can be selected from

(m + n) things.

( )

( )

mn ( )

m

C m n! m n!

m! m n m! m! n!

+ = + = +

+ −

Note 1. If m =n, the groups are equal and in this case the number of different ways of subdivision ( )

2m! 1

=m! m! 1!×

since two groups can be interchanged without getting a new subdivision.

Note 2. If 2m things be divided equally amongst 2 persons, then the number of ways

(2m !).

m! m!

(A) Now (m + n + p) things (m ¹ n ¹ p), to be divided into three groups containing m, n, p things respectively.

m things can be selected out of (m + n + p) things in m+n+pCm ways, then n things out of remaining

(n + p) things in n+pCn ways and lastly p things out of remaining p things in pCp i.e., one way.

Hence the required number of ways is m n p+ + Cm×n p+Cn

( )

( )

m n p n p! m n p !( ) ( )

m! n p! n! p! m! n! p!

= + + × + = + +

+

Note 1. If now m = n = p, the groups are equal and in this case, the different ways of subdivision ( )

3m! 1

=m! m! m! 3!×

since the three groups of subdivision can be arranged in 3! ways.

Note 2. If 3m things are divided equally amongst three persons, the number of ways ( )

3m!

=m! m! m!

SOLVED EXAMPLES

Example 75 : In how many ways can be College Football team of 11 players be selected from 16 players?

Solution :

The required number^1611 ( )

C 16! 16!

= =11! 16 11 !− =11! 5 != 4, 368

Example 76: From a company of 15 men, how many selections of 9 men can be made so as to exclude 3

particular men?

Solution :

Excluding 3 particular men in each case, we are to select 9 men out of (15 – 3) men. Hence the number of

selection is equal to the number of combination of 12 men taken 9 at a time which is equal to

(^12) C 9 12!
= =9! 3 != 220.