Associate Degree 2020 – 2021 First Semester

CCMA4001 Quantitative Analysis I

Exercise 2

1. In a medical study, patients are classified according to whether they have blood type A, B, AB, or O
and also according to whether their blood pressure is low, normal, or high. In how many different
ways can a patient thus be classified according to blood type and blood pressure?

Solution:

4  3  12

2. A four-course dinner consists of a soup, a main dish, a dessert and a drink. If one can select from
3 different soups, 5 main dishes, 4 desserts and 7 drinks, how many dinner choices are possible?

Solution:

3  5  4  7  420

3. In a certain town, each registration number consists of 2 letters (repetition is allowed) followed by
a number between 1 and 9999 inclusive. The letters are limited to those from A to H.
How many registration numbers are possible?

Solution:

8  8  9999  639936

4. Emily has several ways to spend her weekend. She can read a book, watch a video or go for a drink.
She can choose between 7 books, 5 videos and 3 coffee shops. How many choices does she have if
(a) she only takes one activity?
(b) she only takes two activities?
(c) she takes all three sorts of activities?

Solution:

(a) 7  5  3  15

(b) 7  5  7  3  5  3  35  21  15  71

(c) 7  5  3  105
1
5. How many ways are there to arrange 5 balls of different colours in a line?

Solution:

5  4  3  2  1  120 or 5! 120

6. In how many ways can the letters in the word SOCIAL be arranged?

Solution:

6  5  4  3  2  1  720 or 6! 720

7. If 16 entries are submitted to an essay contest, in how many different ways can the judges award a
first prize and a second prize?

Solution:

16  15  240 or P216  240

8. There are 20 candidates for three different marketing jobs. How many different ways to fill the position?

Solution:

20  19  18  6840 or P320  6840

9. A basketball team must schedule a game with each of three different teams. There are five different
dates available for games. How many different schedules can be made?

Solution:

5  4  3  60 or P35  60

2
10. Two different physics books, four different chemistry books, and a biology book are arranged on a shelf.
(a) How many different arrangements are possible?
(b) How many different arrangements are possible if the biology book is put in the first position?
(c) How many different arrangements are possible if only three books are put on the shelf –
a physics book, a chemistry book, and the biology book, in that order?

Solution:

(a) 7! 5040

(b) 6! 720

(c) 2  4  8

11. There are three Chinese books and seven Mathematics books, all of which are different.
(a) In how many ways can arrange 6 of them on a shelf?
(b) In how many ways can arrange the three Chinese books on the left and the seven Mathematics
books on the right of the shelf?
(c) In how many ways can arrange all of them on the shelf with the Chinese books together?

Solution:

(a) P610  151200

(b) 3!7! 6  5040  30240

(c) Treat the three Chinese books as one object, so now eight objects are arranged on the shelf.
And the three Chinese books can be arranged in 3! ways.

Therefore, the number of ways to arrange all the books on the shelf with the Chinese books together is
8!3! 40320  6  241920

3
12. How many four-digit numbers can be made from the numbers 1, 2, 3, 4, 5, 6 and 7, if
(a) repetition is not allowed?
(b) repetition is allowed?

Solution:

(a) 7  6  5  4  840 or P47  840

(b) 7  7  7  7  2401

13. How many different ID cards numbers can be formed if there are six digits on a card and no digit
can be used more than once?

Solution:

10  9  8  7  6  5  151200 or P610  151200

14. How many four-digit numbers can be formed if zero is not an admissible candidate in the
thousandth position and
(a) repetitions of digits are permitted?
(b) repetitions of digits are not permitted?

Solution:

(a) The fist digit can be selected in any one from 1 to 9 and the rest digits can each be selected in
any of 10 ways.

Therefore, the number of four-digit numbers can be formed is

9  10  10  10  9000

(b) There are 9 choices for the thousandth digit because zero is not allowed.
Since we cannot duplicate the first digit, there are still 9 choices available for the hundredth digit
because zero is admissible now; there are 8 choices for the tenth digit and 7 choices for the unit digit.

Therefore, the number of four-digit numbers can be formed is

9  9  8  7  4536

4
15. An automobile license plate consists of three letters followed by four digits.
(a) How many different plates can be made if repetitions are allowed?
(b) How many different plates can be made if repetitions are not allowed?
(c) How many different plates can be made if repetitions are allowed in the letters but not in the digits?

Solution:

(a) 26  26  26  10  10  10  10  175760000

(b) 26  25  24  10  9  8  7  78624000 or P326  P410  15600  5040  78624000

(c) 26  26  26  10  9  8  7  88583040 or 26  26  26  P410  17576  5040  88583040

16. A five-digit number is formed from the digits 1, 2, 3, 4 and 5 and repetition are not allowed.
(a) How many five-digit numbers can be formed?
(b) How many five-digit numbers formed is divisible by 5?
(c) How many five-digit numbers formed is an odd number?

Solution:

(a) The number of five-digit numbers can be formed is 5! 120

(b) The unit digit must be 5, so the number of five-digit numbers formed is divisible by 5 is
4! 24

(c) The unit digit can be chosen from the digits 1, 3 and 5, so the number of five-digit numbers
formed is an odd number is
3  4! 3  24  72

5
17. A three-digit number is formed from the digits 0, 1, 3, 5, 6, 7 and 8. Each digit can be used only once.
How many possible numbers are even numbers?

Solution:

The number of three-digit numbers formed ending with ‘0’ is P26  6  5  30

The number of three-digit numbers formed ending with ‘6’ or ‘8’ and with the tenth digit is ‘0’ is
2  P15  2  5  10

The number of three-digit numbers formed ending with ‘6’ or ‘8’ and does not contain ‘0’ is
2  P25  2  5  4  40

Therefore, the total number of three-digit numbers formed are even numbers is
30 + 10 + 40 = 80

6
18. (a) How many different numbers altogether can be formed by taking one, two, three and four digits
from the digits 9, 8, 3 and 2, repetition not being allowed?
(b) How many of the numbers in part (a) are odd and greater than 800?

Solution:

(a) The number of 4-digit numbers can be formed is P44  4! 24

The number of 3-digit numbers can be formed is P34  4  3  2  24

The number of 2-digit numbers can be formed is P24  4  3  12

The number of 1-digit numbers can be formed is P14  4

Therefore, the number of different numbers altogether can be formed is

P44  P34  P24  P14  24  24  12  4  64

(b) The number of 4-digit numbers formed ending with ‘3’ is P33  3! 6

The number of 4-digit numbers formed ending with ‘9’ is P33  3! 6

The number of 3-digit numbers formed beginning with ‘8’ and ending with ‘3’ is P12  2

The number of 3-digit numbers formed beginning with ‘8’ and ending with ‘9’ is P12  2

The number of 3-digit numbers formed beginning with ‘9’ and ending with ‘3’ is P12  2

Therefore, the number of different numbers can be formed are odd and greater than 800 is
P33  P33  P12  P12  P12  6  6  2  2  2  18

7
19. Four couples take a row of seats at a train.
(a) How many different ways can they be seated?
(b) How many different ways can they be seated if each couple is to sit together?
(c) How many different ways can they be seated if all women are to sit together?
(d) How many different ways can they be seated if all men are to sit together and all women are
to sit together?
(e) How many different ways can they be seated if no two people of the same sex are allowed to
sit together?

Solution:

(a) There are four couples, i.e., 8 people. The number of ways is 8! 40320

(b) Treat each couple as one object, so now four objects are to be arranged in a row.
And each couple can be arranged in 2! ways.

Therefore, the number of different ways they can be seated if each couple is to sit together is
4!2!2!2!2! 24  2  2  2  2  384

(c) Treat the four women as one object, so now five objects are to be arranged in a row.
And the four women can be arranged in 4! ways.

Therefore, the number of different ways they can be seated if all women are to sit together is
5!4! 120  24  2880

(d) Treat the four men as one object, and the four women as one object so now two objects are to
be arranged in a row.
And the four men can be arranged in 4! ways and the four women can be arranged in 4! ways.

Therefore, the number of different ways they can be seated if all men are to sit together
and all women are to sit together is
2!4!4! 2  24  24  1152

(e) The four men can be arranged in 4! ways and the four women can be arranged in 4! ways.
The first seat can be taken by a man or a woman (2 ways) and then arrange them to sit alternatively.

Therefore, the number of different ways they can be seated if no two people of the same sex are
allowed to sit together is
4!4!2  24  24  2  1152

8
20. Three married couples have bought six seats in a row for a performance of a musical comedy.
(a) In how many ways can they be seated?
(b) In how many ways can they be seated if each couple is to sit together with the husband to the
left of his wife?
(c) In how many ways can they be seated if each couple is to sit together?
(d) In how many ways can they be seated if all the men are to sit together and all the women are to
sit together?

Solution:

(a) 6! 720

(b) 3! 6

(c) 3!2!2!2! 6  2  2  2  48

(d) 2!3!3! 2  6  6  72

21. (a) How many different ways can five persons line up to get on a bus?
(b) In how many ways can they line up if two of the persons must follow each other?
(c) In how many ways can they line up if two of the persons refuse to follow each other?

Solution:

(a) 5! 120

(b) 4!2! 24  2  48

(c) 5!4!2! 120  48  72 OR 3! P24  72

22. In how many ways can ten pupils line up if the two youngest pupils are separated?

Solution:

10!9!2! 3628800  725760  2903040 OR 8! P29  2903040

9
23. John and his family (wife, father, mother, 3 sons and 1 daughter) take photograph at the studio.
They are arranged to stand in one row, how many arrangement are there if
(a) there is no restriction;
(b) John’s parents stand in the middle positions;
(c) John’s parents stand neither the leftmost nor rightmost of the row;
(d) the three women stand together;
(e) each pair of the couples stand together;
(f) no two women stand next to one another.

Solution:

(a) 8! 40320

(b) 6!2! 720  2  1440

(c) P26  6! 30  720  21600

(d) 6!3! 720  6  4320

(e) 6!2!2! 720  2  2  2880

(f) 5! P36  14400

10
24. A group of people with 5 men and 4 women take photograph.
(a) They are arranged to stand in one row, how many arrangement are there if
(i) there is no restriction;
(ii) men and women stand alternate;
(iii) there is a couple and they stand together.
(b) They are arranged to stand in two rows, how many arrangement are there if
(i) men and women not stand in the same row;
(ii) there is no restriction for the number of people in each row.

Solution:

(a) (i) 9! 362880

(ii) 4!5! 24  120  2880

(iii) 8!2! 40320  2  80640

(b) (i) 4!5!2! 24  120  2  5760

(ii) 9!8  362880  8  2903040

OR P19  8! P29  7! P39  6! P49  5! P59  4! P69  3! P79  2! P89 1!
 (9  40320  72  5040  504  720  3024 120)  2
 2903040

11
25. Six chairs are placed in a straight line. Find the number of different ways in which 6 students A, B,
C, D, E and F may sit on these chairs so that
(a) A and B sit next to each other;
(b) C and D do not sit next to each other;
(c) A and B sit next to each other and C and D sit next to each other;
(d) A and B sit next to each other and C and D do not sit next to each other.

Solution:

(a) 5!2! 120  2  240

(b) 6!5!2! 720  240  480

(c) 4!2!2! 24  2  2  96

(d) 5!2!4!2!2! 240  96  144

12
26. Twelve people (6 men and 6 women) are arranged along a row of 12 seats. Among them, there are
two couples, Peter and Karen, Leon and Amy.
(a) How many possible arrangements are there if there is no restriction?
(b) How many possible arrangements if men and women sit alternately?
(c) How many possible arrangements are there if four men should be seated in the middle four seats?
(d) How many possible arrangements are there if Peter and Karen are sitting together?
(e) How many possible arrangements are there if Peter, Karen, Leon and Amy are sitting together?
(f) How many possible arrangements are there if Peter and Karen are sitting together while Leon
and Amy are separated?

Solution:

(a) 12!  479001600

(b) 6! 6! 2  1036800

(c) P46  8!  14515200

(d) 11! 2!  79833600

(e) 9! 4!  8709120

(f) n(Peter & Karen together) – n(Peter & Karen together, Leon & Amy together)
 11! 2! 10! 2! 2!  65318400

13
27. Ten students attend a meeting and are arranged along a row of 10 seats. Among them, there are
David, John and Kelly.
(a) How many different arrangements are there if David and John sit in the fifth and ninth position
respectively?
(b) How many different arrangements are there if John and Kelly sit at either the leftmost or
rightmost position?
(c) How many different arrangements are there if David is the first one?
(d) How many different arrangements are there if David is not the first one?
(e) How many different arrangements are there if David is not the first one and John is not the
second one?
(f) How many different arrangements are there if David is not the first one, John is not the second
one and Kelly is the third one?

Solution:

(a) 8!  40320

(b) 8! 2  80640

(c) 9!  362880

(d) 10! 9!  3265920 OR 9  9!  3265920

(e) (10! 9!)  (9! 8!)  2943360 OR 9  9! 8  8!  2943360

(f) (9! 8!)  (8! 7!)  287280 OR 8  8! 7  7!  287280

14
28. The back row of a lecture hall has 12 seats, all of which are empty. A group of 8 students, including
Martin and Roger, sit in this row. Find the number of different ways they can sit in these 12 seats if
(a) there are no restrictions;
(b) Martin and Roger do not sit in seats which are next to each other;
(c) all 8 students sit together with no empty seat between them.

Solution:

(a) P812  19958400

(b) Number of different ways Martin and Roger sit in seats which are next to each other is
P711  2! 1663200  2  3326400

Therefore, number of different ways Martin and Roger do not sit in seats which are next to
each other is
P812  P711  2! 19958400  3326400  16632000

(c) 8!5  40320  5  201600

29. How many possible ways can we arrange the letters from A to J so that no two vowels occur consecutively?

Solution:

There are seven consonants B, C, D, F, G, H and J and three vowels A, E and I.

Imagine that the 7 different consonants are arranged in a row and we determine the number of
possible ways of inserting the 3 different vowels in the 8 spaces before, between, and after the
consonants such that no 2 vowels are placed next to each other.

Number of ways to arrange the 7 consonants in a row is 7! 5040

And for a particular arrangement of the 7 consonants, the number of ways of placing the 3 vowels is
P38  8  7  6  336

Therefore, the number of ways we can arrange the letters from A to J

so that no two vowels occur consecutively is
7!P38  5040  336  1693440

15
30. In how many ways can the letters of the word STATISTICS be arranged?

Solution:

There are ten letters and S occurs three times, T occurs three times, I occurs twice.

10!
Therefore the number of ways is  50400
3!3!2!

31. A hostel has 3 bedrooms and there are 13 girls. One bedroom has 6 beds, the second has 4 beds,
and the third has 3 beds. In how many different ways can the girls be assigned rooms?

Solution:

13!
 60060
6!4!3!

32. (a) How many ways can the letters in the word MINIMUM be arranged?
(b) How many ways with all the three M’s arranged together?
(c) How many arrangements begin with MMM?

Solution:

(a) There are seven letters with M occurs three times and I occurs twice.

7!
Therefore the number of ways is  420
3!2!

(b) Treat the three M’s as one object, so there are five objects with I occurs twice.
And there is only one way of arranging MMM.

5!
Therefore the number of ways is  60
2!

(c) The arrangements must begin with MMM, so only the remaining four letters are to be arranged
with I occurs twice.
And there is only one way of arranging MMM.

4!
Therefore the number of ways is  12
2!
16
33. A shop window designer has 7 balloons, of which 1 is white, 2 are blue and 4 are red.
She hangs these balloons in a line in the shop front.
(a) Find the number of arrangements she can make by using all 7 balloons.
(b) Find the number of arrangements she can make by using only 6 balloons.

Solution:

(a) There are seven balloons of which 1 is white, 2 are blue and 4 are red.

7!
Therefore the number of arrangements is  105
2!4!

(b) If only six balloons are used, the balloons can be 2 blue and 4 red; 1 white, 1 blue and 4 red;
or 1 white, 2 blue and 3 red.

6!
The number of ways to arrange 2 blue balloons and 4 red balloons is  15
2!4!
6!
The number of ways to arrange 1 white balloon, 1 blue balloon and 4 red balloons is  30
4!
6!
The number of ways to arrange 1 white balloon, 2 blue balloons and 3 red balloons is  60
2!3!

Therefore, the number of ways to arrange six balloons is

6! 6! 6!
   15  30  60  105
2!4! 4! 2!3!

17
34. Suppose there are 10 identical balls. Find the number of ways to put these balls into 4 different bags.
Notice that empty bags and bags with more than one ball is allowed.

Solution:

We can use partition method.

Assume x is a ball and | is a separator for bags

For example,

xxx|x|xxxx|xx represents three, one, four and two balls are put in the first, second, third and
fourth bags respectively.

xxxxxxxxxx| | | represents all ten balls are put in the first bag.

xxxxx| |xxx|xx represents five, three and two balls are put in the first, third and fourth bags respectively
and the second bag is empty.

So the number of ways to put 10 balls into 4 different bags is equivalent to the number of ways to
arrange the ten x’s and three bars.

(10  3)! 13!

Therefore, the number of ways is   286
10!3! 10!3!

35. Suppose there are 15 identical balls. Find the number of ways to put these balls into 7 different bags.
Notice that empty bags and bags with more than one ball is allowed.

Solution:

(15  6)! 21!

  54264
15!6! 15!6!

18
36. How many ways can a window dresser display four different shirts in a circular arrangement?

Solution:

(4  1)! 3! 6

37. Six bulbs are planted in a ring and two do not grow. How many ways if the two that do not grow are
next to each other?

Solution:

Treat the two bulbs that do not grow as one object and so now five objects are to be arranged in a ring.
And the two bulbs that do not grow can be arranged in 2! ways.

Therefore, the number of ways if the two bulbs that do not grow are next to each other is
(5  1)!2! 4!2! 24  2  48

38. A family (Dad, Mum and two children) attends a wedding banquet of their relatives. In how many
different ways can they sit at a round table of 12 persons, if all the family members are to sit together?

Solution:

(9  1)!4! 8!4! 40320  24  967680

39. A family of 7 has dinner at a round table. Find the number of ways they can be seated if
(a) there is no restriction,
(b) the parents must be seated together,
(c) the parents must not be seated together.

Solution:

(a) (7  1)! 6! 720

(b) (6  1)!2! 5!2! 120  2  240

(c) 720  240  480

19
40. Four couples are arranged to sit at a round table. How many arrangements can be made so that
(a) a particular couple is to sit together?
(b) all couples are to sit together?
(c) no two women are allowed to sit together?

Solution:

(a) (7  1)!2! 6!2! 720  2  1440

(b) (4  1)!2!2!2!2! 3!2!2!2!2! 6  2  2  2  2  96

(c) First, the number of ways to arrange the four men to sit at a round table is (4  1)! 3! 6

Then the women have to sit in, one between two men, so that no two women are sit together.

Since M 1 is fixed, the number of ways to arrange the four women is 4! 24

Therefore, the number of ways to arrange the four couple to sit at a round table
so that no two women are allowed to sit together is
(4  1)!4! 3!4! 6  24  144

41. In how many ways can 5 students will be chosen from a class of 20 and arranged to sit in a round table?

Solution:

C 520  (5  1)! 15504  24  372096

42. Find the number of ways in which 10 persons to be seated at two round tables, each having 5 seats.

Solution:

5  (5  1)!(5  1)!  252  24  24  145152

C10

20
43. (a) In how many ways can one white, one blue, one red and two yellow beads be threaded on a ring
to make a bracelet?
(b) Find the number of ways that the red and white beads are next to each other.

Solution:

(a) If all the beads are different colors, the number of ways of arranging five beads on a ring is
(5  1)! 4!
  12 ,
2 2
(5  1)!
but since there are two yellows, the number of ways is  2! 12  2  6
2

(b) Treat the red and white beads as one object and the red and white beads can be arranged in 2! ways.

The number of ways that the red and white beads are next to each other is
 (4  1)! 
 2  2!  2! 3  2  2  3
 

44. A college catalogue lists 15 elective courses for the students. Find the number of ways in which a
student can select
(a) 4 courses;
(b) 3 courses;
(c) 2 courses;
(d) 1 course.

Solution:

15! 15!
4 
(a) C15   1365
(15  4)!4! 11!4!

15! 15!
3 
(b) C15   455
(15  3)!3! 12!3!

15! 15!
2 
(c) C15   105
(15  2)!2! 13!2!

15! 15!
1 
(d) C15   15
(15  1)!1! 14!1!

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45. In how many different ways can a student select two of six mathematics courses together with
three of seven English courses?

Solution:

The number of ways to select two mathematics courses is C 62  15

and the number of ways to select three English courses is C73  35

Therefore, the number of ways to select two mathematics courses and three English courses is
C 62  C 37  15  35  525

46. A student committee must consist of two juniors and four seniors. If seven juniors and eight seniors
are willing to serve on the committee, in how many different ways can it be selected?

Solution:

C 72  C 84  21  70  1470

47. (a) In how many ways can 5 hearts be drawn from an ordinary pack of 52 playing cards?
(b) In how many ways can 3 hearts and 2 diamonds be drawn?

Solution:

5  1287
(a) C13

3  C 2  286  78  22308
(b) C13 13

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48. A carton of 12 transistor batteries contains 4 that are defective. Find the number of different ways
can one choose 3 of these batteries so that
(a) none of the defective batteries is included;
(b) exactly 1 of the defective batteries is included;
(c) exactly 2 of the defective batteries are included;
(d) exactly 3 of the defective batteries are included.

Solution:

(a) C 83  56

(b) C 82  C14  28  4  112

(c) C18  C 42  8  6  48

(d) C 34  4

49. A small cake shop offers six different kinds of cake every day: strawberry, blueberry, chocolate,
mango, lemon and New York cheese in the following quantity. There are 8 pieces of strawberry
cake, 7 pieces of blueberry cake, 9 pieces of chocolate cake, 8 pieces of mango cake, 7 pieces of
lemon cake and 7 pieces of New York cheese cake. All cakes are treated as distinct objects even if
they are of the same kind. Find the number of ways on choosing seven cakes if
(a) all are of different kinds;
(b) all are of the same kind;
(c) there is no lemon cake and mango cake;
(d) more than half are chocolate cakes.

Solution:

(a) 0

(b) C87  C77  C97  C87  C77  C77  55

7  2629575
(c) C31

(d) C94  C37

3  C5  C 2  C6  C1  C 7  1066080
9 37 9 37 9

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50. For each of the following examination papers, determine the number of ways a candidate may
choose the required number of questions. Ignore the order in which questions are attempted.
(a) The paper has 2 sections, A and B. A has 6 questions from which candidates must answer any 4;
B has 5 questions from which candidates must answer any 3.
(b) The paper has 5 questions and the candidate must answer at least 3 questions.
(c) The paper has 3 sections, A, B, and C, each with 2 questions. Candidates must answer 4 questions
in total, including at least 1 from each section.

Solution:

(a) C 64  C53  15  10  150

(b) C53  C54  C 55  10  5  1  16

(c) The candidate has C13 ways to choose a section to answer 2 questions.
Therefore, the number of ways the candidates must answer 4 questions in total,
including at least 1 from each section is
C13  C 22  C12  C12  3  1  2  2  12

51. In a library, 9 of 10 different books are selected and arranged on a bookshelf. Find the number of
arrangement if
(a) there are no restrictions;
(b) they are in the order of the library serial number;
(c) the books with the smallest and the largest serial numbers among the 10 books are put on the
left and right ends respectively;
(d) the books with the smallest and the largest serial numbers among the 9 selected books are put
on the left and right ends respectively.

Solution:

(a) P910  3628800

(b) The library serial numbers can be in ascending or descending order.

 The required number of arrangements is 2  C10

9  20

(c) P78  40320

(d) There are C109 ways to select 9 books.

Among the 9 selected books, except those with the smallest and the largest serial numbers,
there are 7! ways to arrange other 7 books.
 The required number of arrangements is C109  7!  50400

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52. There are two boxes, A and B. In box A, it contains 3 blue balls and 5 yellow balls.
In box B, it contains 4 blue balls and 4 yellow balls. The balls’ sizes are identical.
(a) How many different patterns of colour can be formed when the eight balls in box A are arranged
in a row?
(b) If three balls are drawn from box A, find the number of ways that
(i) the three balls have the same colour;
(ii) the number of yellow balls is more than the number of blue balls.
(c) Two balls are drawn from each of the boxes A and B. Find the number of ways that the numbers
of yellow balls drawn from each box are the same.

Solution:

8!
(a)  56
3!5!

(b) (i) C 33  C 50  C 30  C 53  1  10  11

(ii) C13  C 52  C 30  C 53  30  10  40

(c) If the numbers of yellow balls drawn from each box are the same, then there must be 2 blue balls,
1 blue ball and 1 yellow ball or 2 yellow balls drawn from each box.

Therefore, the number of ways that the numbers of yellow balls drawn from each box are the same is
C 32  C 50  C 42  C 04  C13  C15  C14  C14  C 30  C 52  C 04  C 42  18  240  60  318

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53. A box contains 30 rabbits, five of which are male and the rest of which are female.
(a) How many ways can 2 male rabbits and 3 female rabbits be chosen from the box?
(b) How many ways can 2 male rabbits and 3 female rabbits be chosen if the first choice has to be
a female rabbit and the two kinds must be chosen alternatively?

Solution:

(a) We are considering the number of possible ways in which 2 male rabbits can be chosen from 5
and 3 female rabbits can be chosen from 25.
The sequence of choosing the rabbits is unimportant, so this is combination.
By the principle of multiplication, the total number of ways is
C 52  C 325  10  2300  23000

(b) Under this circumstances, the order of choice is crucial, so we can no longer use combination.
The only order that can happen based on the above condition is
(1) female, (2) male, (3) female, (4) male, (5) female

There are 25 female rabbits from which the first rabbit can be chosen.
There are 5 male rabbits from which the second can be chosen.
There are 24 female rabbits from which left for the third selection,
4 male rabbits for the fourth selection and 23 female rabbits for the last selection.

Thus, by the principle of multiplication, the total number of ways is

25  5  24  4  23  276000

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54. A team of four is chosen from six boys and five girls.
(a) In how many ways can the team be chosen if there are no restrictions?
(b) In how many ways can the team be chosen if the team does not contain boy?
(c) In how many ways can the team be chosen if the team contains only one boy?
(d) In how many ways can the team be chosen if the team contains at least one boy?
(e) In how many ways can the team be chosen if there must be more boys than girls?

Solution:

(a) There are 11 people, from whom four are chosen. The order in which they are chosen is not important.
The number of ways is C11 4  330

(b) The number of ways of choosing the team does not contain boy
= The number of ways of choosing four girls
 C54
5

(c) The number of ways of choosing one boy and three girls is C16  C 53  6  10  60

(d) If there are at least one boy, then there must be one boy and three girls; two boys and two girls;
three boys and one girl or four boys.

The number of ways of choosing one boy and three girls is C16  C 53  6  10  60

The number of ways of choosing two boys and two girls is C 62  C 52  15  10  150
The number of ways of choosing three boys and one girl is C 36  C15  20  5  100

The number of ways of choosing four boys is C 64  15

Therefore, the number of ways of choosing the team contains at least one boy is
C16  C 53  C 62  C 52  C 63  C15  C 64  60  150  100 `15  325

OR The number of ways of choosing the team contains at least one boy
= The number of ways of choosing the team of four
– The number of ways of choosing the team does not contain boy
 C11
4  C4
5

 330  5
 325

(e) If there are more boys than girls, then there must be three boys and one girl or four boys.

Therefore, the number of ways the team can be chosen if there must be more boys than girls is
C 36  C15  C 64  100  15  115
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55. There are five men and seven women in a department.
(a) How many ways can a committee of four people be selected?
(b) How many ways can this committee be selected if all four people selected are men?
(c) How many ways can this committee be selected if there must be three men and one woman in
the committee?
(d) How many ways can this committee be selected if there must be two men and two women in
the committee?
(e) How many ways can this committee be selected if there must be at least two women in the
committee?

Solution:

4  495
(a) C12

(b) C 54  5

(c) C 53  C17  10  7  70

(d) C 52  C 72  10  21  210

(e) C 52  C 72  C15  C 37  C 74  10  21  5  35  35  210  175  35  420

4  C 4  C 3  C1  495  5  10  7  495  5  70  420

OR C12 5 5 7

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56. A committee of eight members consists of one married couple together with four other men and
two other women. From the committee, a working party of five persons to be formed.
(a) Find the number of different working parties which can be formed.
(b) Find the number of different working parties if it must contain both the husband and his wife.
(c) Find the number of different working parties if it must not contain both the husband and his wife.
(d) Find the number of different working parties if it must contain three men and two women.
(e) Find the number of different working parties if it must contain at least one man and at least one woman.

Solution:

(a) C 85  56

(b) C 36  20

(c) C 85  C 36  56  20  36

(d) C 53  C 32  10  3  30

(e) C 54  C13  C 53  C 32  C 52  C 33  5  3  10  3  10  1  15  30  10  55

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57. Mary has 12 friends and hope to invite 6 of them to a party. Find the number of ways she may do this if
(a) there is no restriction on choice;
(b) two of the friends is a couple and will have the same choices;
(c) two of the friends will not attend together.

Solution:

6  924
(a) C12

(b) The couple will have the same choices means they will attend or reject together.

If the couple attends, the remaining 4 may then be chosen from the other 10.
Therefore, the number of ways is C10 4  210

If the couple does not attend, Mary simply chooses 6 from the other 10.
Therefore, the number of ways is C10 6  210

Hence, the number of ways if two of the friends is a couple and will have the same choices is
4  C 6  210  210  420
C10 10

(c) Let the two friends be A and B.

If one of A or B attends, the number of ways the party can be formed is

C12  C10
5  2  252  504

If neither A nor B attends, the number of ways the party can be formed is
C106  210

Hence, the number of ways if two of the friends will not attend together is
C12  C10
5  C 6  504  210  714
10

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58. A committee consists of a president, a treasurer and a secretary. The three posts are to be chosen
from a club consisting of 10 people (A, B, C, D, E, F, G, H, I and J). Find the number of ways to
form the committee if
(a) there is no restriction;
(b) A will serve;
(c) B will serve only if he is president;
(d) C and D will serve together or not at all;
(e) E and F will not serve together.

Solution:

(a) P310  720

(b) C92  3!  216

(c) P29  P39  576

(d) C18  3! P38  384

(e) C82  3! 2  P38  672

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59. A committee consists of a chairman, a treasurer and a secretary. The three posts are to be chosen
from class A and class B. There are 15 students in class A and 20 students in class B respectively.
Find the number of ways to form the committee if
(a) there is no restriction;
(b) the chairman is from class A;
(c) not all members are of the same class;
(d) the secretary and the treasurer are not of the same class.

Solution:

(a) P335  39270

(b) 15  P234  16830

(c) P335   P315  P320   29700

(d) 15  20  33  2  19800

60. There are 5 classes of Form 6 in a secondary school. To form a task group of 20 members,
4 representatives are nominated by each class. From the task group, 5 members are randomly selected.
Find the number of ways to select the 5 members if they are nominated by
(a) five different classes;
(b) four different classes;
(c) three different classes.

Solution:

(a) (C14 )5  1024

(b) C54C14C 42 (C14 )3  7680

(c) C53 C32 (C42 )2 C14  C13C34 (C14 ) 2   6240

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