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1.0 Introduction 2.0 Objectives 3.0 Main Content

3.1 Definition of Combination 3.2 Evaluation of n Combination r 4.0 Conclusion

5.0 Summary

6.0 Tutor-Marked Assignment 7.0 References/Further Reading

1.0 INTRODUCTION

Suppose we have 5 cards labelled A, B, C, D, and E, then we can arrange the cards in threes i.e. ⁵P3 = 60 ways. Since the order of arrangement is relevant and important, then ABC, BAC and CAB are two different arrangements. Now if the order of arrangement is irrelevant and less important then all arrangements such as ABC BAC, CAB, ACB, etc are all the same and shall be counted as one type of arrangement. This type of arrangement in which the order of arrangement is less important is called Combination.

2.0 OBJECTIVES

At the end of the unit, you should be able to:

 evaluate combinational expressions

 recognise a problem of selection or combination.

3.0

MAIN CONTENT

3.1

Definition of Combination

While permutation is about the arrangement of objects taking r (r ≤ n) objects at a time (with interest in order of arrangement), combination is about selecting a certain number of r objects from a set of n objects, in this case, the order of arrangement is of no interest nor importance, thus if there are n numbers of objects, then the number of ways of selecting r objects is defined as:

nCr=

20P₅ nCris also denoted by: ⁿCror

3.2

Evaluation of n Combination r

Evaluate the following;

(a) 20C15 (b) ²⁰C5÷²⁰P5 (c) IfⁿP3= 4(ⁿC5) find the value of n.

Solution

1. ²⁰C15 = =

2. 20C5÷ ÷ =

3. nP3= 4ⁿC5

4(n - 3) (n - 4) = 5! = 120 (n - 3) (n - 4) = 30

n²- 7n + 12–30 = 0

n²–7n–18 = 0 (n - 9 )(n + 2) = 0 n–9 = 0 or n + 2 = 0

n = 9 or n = - 2

But n cannot be negative so n = 9 is the solution.

3.3

Combination or Selection Example 2

1. In how many ways can 6 questions be chosen from 10 optional questions?

Solution

No. of selections is ¹⁰C6 =

ways

2. A committee of 3 men and 2 women is to be chosen from 7 men and 5 women. In how many ways can the selection be made?

Solution

No. of selection is⁷C3AND⁵C2=⁷C3

5C2= 35 10 = 350ways

3. A box contains 12 male voter’s cards and another box contains 10 female voter’s cards. Find the number of ways of selecting 5male cards and 5 female cards if 2 particular female cards must be selected.

Solution

Selecting male cards =¹²C5= 792ways

Selecting the remaining 3 female cards = ⁸C3 = 56ways Selection = 792 56 = 44352ways

4. A committee of 6 men and 4 women is to be formed from 10 men and 8 women. (a) In how many ways can this committee be formed? (b) If a particular woman will not serve on the committee with a particular man, find the number of ways this committee can be formed.

Solution

(a) No. of committee =¹⁰C6

8C4= 210 70 = 14700

(b) No. of committee with a particular woman not serving with a particular man = (no. of possible committee) - (no. of committee involving the two of them) = 14700 - (⁹C5

7

C3)

= 14700 - 4410 = 10290ways

5. Find the number of ways of choosing four party agents from five men and three women if at least one woman must be in the selection.

Solution

At least one woman = (1W AND 3M) OR (2W AND 2M) OR (3W AND 1M)

= (³C1

5C3) + (³C2

5C2) + (³C3

5C1)

= 30 + 30 + 5 = 65ways

6. An opposition group of five is to be formed from five men and four women. Calculate the number of ways this can be done if the number of men must be greater than the number of women.

Solution

For number of men greater than number of women

= (5M AND 0W) OR (4M AND 1W) OR (3M AND 2W)

= (⁵C5

4C0) + (⁵C4

4C1) + (⁵C3

4C2)

= 1 + 20 + 60 = 81 ways.

SELF-ASSESSMENT EXERCISE

i. Five delegates out of five men and five women are to represent an electoral body at a national debate. Find the number of ways the delegates can be selected if (a) it must consist of 2 men and 3 women (b) the delegates must have at least one person of each sex.

ii. In how many ways can 5 secretaries, 4 clerks, 3 drivers and 2 cleaners be chosen from 10 secretaries, 9 clerks, 8 drivers and 7 cleaners?

4.0 CONCLUSION

In this unit, you have learnt how to calculate the number of ways a given number of objects can be arranged such that order of arrangement is irrelevant.

5.0 SUMMARY

From this unit we learnt that;

Combination means selection or choosing r distinct objects from n

distinct objects.

nCr= ,ⁿC0= 1 and ⁿC1= n

nCr=

6.0 TUTOR-MARKED ASSIGNMENT

1. The INEC Chairperson is to select 2 regional coordinators for each of the 6 political zones in Nigeria. If there are 15 qualified people for these post, calculate the number of ways he can do the selection, (assuming any person can be taken to any region).

2. In a country there were 30 political parties. Later a concession was

reached to form two mega parties of at least 13 political parties in each mega party. In how many ways can this be done assuming all the 30 parties must belong to any of the two mega parties?

3. A caretaker committee of five is to be selected from six men and four women to run the affair of a local government. How many different committees can be formed if (a) any person can be taken (b) the number of men on the committee must be greater than the number of women?

7.0

REFERENCES/FURTHER READING

Adekola, A.O. (1993). Further Mathematics. Nigeria: Macmillan Nigeria Publishers Ltd.

Goodman, A. & Talbert, J.F. (1999). Additional Mathematics for West Africa. Longman Group Ltd.

Tuttuh - Adegun, M.R; Sivasubramaniam, S. and Adegoke, R. (2004).

Further Mathematics Project. Ibadan, Nigeria: NPS Educational Publishers Limited.

Stroud, K.A. (1995). Engineering Mathematics. (4^th ed.). Macmillan Press Ltd.

MODULE 4

Unit 1 Nature and Importance of Statistical Inquiries Unit 2 Basic Research Methodology I

Unit 3 Basic Research Methodology II \ Unit 4 Nature of Science

Unit 5 Some Basic Concepts in Social Statistics

UNIT 1 NATURE AND IMPORTANCE OF STATISTICAL