Paper 4: Fundamentals of Business Mathematics & Statistic

2.48 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Algebra

Case (iv) can be selected in^4 C 4 × 7 C 1 ways

Hence the total number of selections, in each case of which at least one lady is included

=^4 C 1 × 7 C 4 +^4 C 2 × 7 C 3 +^4 C 3 × 7 C 2 +^4 C 4 × 7 C 1

= 4 × 35 + 6 × 35 + 4 × 21 + 1 × 7

= 140 + 210 + 84 + 7 = 441.

Example 80: In how many ways can a boy invite one or more of 5 friends?

Solution :

The number of ways =^5 C 1 +^5 C 2 +^5 C 3 +^5 C 4 +^5 C 5 = 2^5 – 1 = 32 – 1 = 31.

Example 81 : In a group of 13 workers contains 5 women, in how many ways can a subgroup of 10 workers

be selected so as to include at least 6 men? [ICWA (F) Dec 2005]

Solution :

In the given group there are 8 (= 13 – 5) men and 5 women in all. Possible cases of forming the subgroup of

10 workers.

men women selections

(i) 6 4 8 C 6 × 5 C 4 = 28 × 5 = 140

(ii) 7 3 8 C 7 × 5 C 3 = 8 × 10 = 80

(iii) 8 2 8 C 8 × 5 C 2 = 1 × 10 = 10

∴ reqd. no of ways = 230.

Example 82 : In how many ways 15 things be divided into three groups of 4, 5, 6 things respectively.

Solution :

The first group can be selected in^15 C 4 ways :

The second group can be selected in (15 4−)C C 5 =^115 ways ;

and lastly the third group in^6 C 6 = 1 way.

Hence the total number of ways =^15 C 4 × 11 C 5

= 15! 11! 15!× =

4!11! 5!6! 4! 5! 6!

Example 83 : A student is to answer 8 out of 10 questions on an examination :

(i) How many choice has he?

(ii) How many if he must answer the first three questions?

(iii) How many if he must answer at least four of the first five questions?

Solution :

(i) The 8 questions out of 10 questions may be answered in^10 C 8

Now^108 ( )

C 10! 10 9 8! 5 9 45

8!2! 8! 2!

= = × × = × =

ways

(ii) The first 3 questions are to be answered. So there are remaining 5 (= 8 – 3) questions to be answered out

of remaining 7 ( = 10 – 3) questions which may be selected in^7 C 5 ways.