COMBINATION

SUBJECT: FURTHER MATHEMATICS

CLASS: SSS 1

 

WEEK FOUR

TOPIC: COMBINATION

SUB-TOPICS:


  1. Combination (selection).
  2. Conditional arrangements and selection.
  3. Probability problems involving arrangement and selection.

SUB-TOPIC 1

Combination (Selection)

In many situations, we make selection without regard to the order. If a committee of 4 members is to be formed from 7 members of staff of DLHS, the order in which the numbers of the given committee are selected is not important.

Combination is therefore a way of selecting items from a collection such that (unlike permutation) the order of selection does not matter.

In selecting three colours from 5 colours: (B, G, R, W, Y), BGR, BRG, GBR, RBG, RGB, are counted as 6 different arrangements (permutations), although they consist of the same 3 colours. The 6 permutations thus represent one combination. Thus, each combination of three objects yields 3! permutation.

Now, the number of the permutations of 5 colours taking 3 at a time, i.e,

The number of combinations of 5 colours taking 3 at a time, i.e,

In general,

Examples

  1. Out of the five science club members of a school, A, B, C, D and E, just three are to be chosen to represent the school in an exhibition. In how many ways can this be done?
  2. In how many ways can a committee of 3 chemistry teachers and 5 mathematics teachers be formed from 6 chemistry teachers and 10 mathematics teachers?

Solution

  1. The three representatives can be selected in
  2. The chemistry teacher can be selected in 6C3 ways and the mathematics teacher can be selected in 10C5 ways.

Total number of ways = 6C3 x 10C5

=

=

=

=

Class activity

  1. In how many ways can a disciplinary committee of 3 be formed from 10 members of staff of a college?
  2. A man travels from Nigeria to Ghana by air and from Ghana to Liberia by ship. He returns by the same means. He has 6 airlines and 4 shipping lines to choose from. In how many ways can he make his journey without using the same airline or shipping line twice?

SUB-TOPIC 2

Conditional arrangement and selection

When restrictions are placed on arrangements or selection, then, the permutation or combination is said to be conditional.

Example 1:

Find the number of ways 6 people can be seated in a round table if two particular friends must sit next to each other.

Solution:

If two people must seat next to each other, the number of ways these friends can sit is 2!

Therefore, the number of ways six people can sit in a round table with two friends that must be together is

Example 2:

A committee of 4 people is to be chosen from 5 married couples. Find how many ways the committee can be chosen if: (i) everyone is equally eligible; (ii) the committee should include at least one woman.

Solution:

i) 5 married couples includes 5 men and 5 women. Since everyone is equally eligible, then, the possible ways of selecting 4 people for the committee are:

4 men and 0 women or

3 men and 1 women or

2 men and 2 women or

1 man and 3 women or

0 men and 4 women.

i.e,

 

 

ii) If at least one woman must be in the committee, then, the possible ways of selecting 4 members of the committee from the couples (5men & 5women) are:

3 men and 1 woman or 2 men and 2 women or 1 man and 3 women or 4 women

i.e,

 

Class activity

  1. An excursion group of 4 is to be drawn from among 5 boys and 6 girls. Find the number of ways of choosing the excursion group if the group:
  2. is to be made up of an equal number of boys and girls;
  3. is to be either all boys or all girls;
  4. has no restrictions on its composition.
  5. A candidate is expected to attempt 12 out of 15 questions. In how many ways can this be done if:
  6. the candidate is to attempt any 12 question;
  7. the first 8 questions are compulsory;
  8. a question is outside the syllabus and hence cannot be completed.

SUB-TOPICS 3

Probability problems involving arrangement and selection

Example 1:

A box contains 10 red, 3 blue and 7 black balls. If three balls are drawn at random, what is the probability that: (a) all 3 are red, (b) all 3 are blue, (c) one of each colour is drawn?

Solution:

Number of ways of selecting any 3 balls from 20 balls = number of element in the sample space.

Number of ways of selecting 3 red balls out of 10 =

(a) P(all the 3 balls are red) =

(b) P(all 3 are blue) =

(c) P(1 red, 1 blue & 1 black) =

Example 2:

Three-digit numbers are formed from the digits 1, 2, 4, 5 and 6. If repetition is not allowed and a number is picked at random, find the probability that it is a multiple of 5 or an odd number.

Solution:

Since probability is involved, we find the sample space for 3-digit numbers formed from 5-digits i.e 5P3.

n(sample space) =

  1. To get multiple of 5, the last digit must be 5. That implies that

The 1st digit can be any of the remaining 4 digits.

The 2nd digit can be any of the remaining 3 digits

No. of ways = 1 x 4 x 3 = 12 ways.

Pr(multiple of 5) =

  1. To get odd number, the last digit will be any of 1 and 5

The first digit can be any of the remaining four

The second digit will be any of the remaining three.

No. of ways = 2 x 4 x 3 = 24ways.

Pr(odd number) =

Hence, pr(multiple of 5 or odd number) =

=

=

Class activity

  1. How many committee of size 5 consisting of three men and 2 women can be selected from 8 men and 6 women if a certain man must not be in the committee?
  2. A bag contains 5 white, 2 black and 3 green balls. If three ball are drawn at random, find the probability that:
  1. All three are green
  2. All three are white
  3. 2 are white and 1 is black
  4. At least, one is black
  5. 1 of each colour is drawn.

PRACTICE QUESTION

  1. How many committee of size 5 consisting of 3 men and 2 women can be selected from 8 men and 6 women if a certain man must not be in the committee? (a) 315 (b) 525 (c) 840 (d) 1287
  2. In how many ways can 9 bulbs be selected from 4 red, 5 green and 6 yellow bulbs if 3 of each colour are to be selected? (a) 800 (b) 120 (c) 40 (d) 27
  3. The number of ways of arranging 9 men and 8 women in a row, when the women occupy the even places is — (a) (b) (c)!8! (d)
  4. A panel consists of 5 men and 4 women. What is the probability of 4 men and 2 women?

(a) (b) (c) (d) .

5. Five digit numbers are formed from digits 4, 5, 6, 7 & 8

  1. How many of such numbers can be formed if repetition of digit is (i) allowed (ii) not allowed?
  2. How many of the numbers are odd if repetition of digits is not allowed?

EVALUATION

  1. Find the number of ways of arranging 9 men and 8 women in a row, if the women occupy the even places.
  2. If find the value of n.
  3. A panel of 5 jurist is to be chosen from a group of 6men and 7 women. Find the number of different panels that could be formed if: (a) a particular man must serve on the panel (b) there is no restriction.
  4. A business man intends to give a dinner party for 6 of his 10 friends. If 2 of them will not attend the party together, in hoe many ways can he select his guests?
  5. A family of 7 is to be seated round a table. In how many ways can this be done, if the father and the mother are to sit together?
  6. Four delegates are to be chosen from 8 members of staff of a college. If 2 of them are senior members of staff, how many different delegations are possible if: (i) only one of the senior members of staff must be in the delegation? (b) the two senior member of staff must be included?
(Visited 7 times, 1 visits today)

error: Alert: Content is protected !!