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1 (a)^ An ordinary deck of 52 cards consists of four suits (clubs, diamonds, hearts and
spades) with 13 denominations each (ace, 2 to 10, jack, queen, king). Find the
number of selection of 5 cards that are all of the same suit. [2]
(b) In how many ways can we select two books from among five distinct Economics
books, three distinct Physics books and two distinct Art books such that these two
books are from different subjects? [2]
(c) In how many ways can eight distinct books be divided among three students if
Andy gets four books, Belle and Calvin each get two books? [2]
[modified 2009/MI/Prelim/II/7]
[Solution]
(a) Number of ways =
4 13
1 5
C C = 5148
4
1
C : Choose 1 suit from 4 suits
13
5
C : Choose 1 card from the 13 cards
(b) Number of ways
= n an Econs bk & a Phys bk n an Econs bk & an Art bk
+ n a Phys bk & an Art bk
= 5×3 + 5×2 + 3×2 = 31
(c) Number of ways =
8 4 2
4 2 2
C C C or (^4)! 2! 2!
8!
(^2) A group of 12 people consists of 6 married couples.
(i) The 12 people are to be seated randomly at a round table. Find the number of ways
in which the 12 people can be arranged if each married couple is seated together.
[2]
(ii) The group is going on a flight and is assigned to sit in three distinct rows of four
seats each. Find the number of ways in which the 12 people can be arranged if
each row has at least 1 woman. [5]
[2015/VJC/Prelim/II/6]
2 [Solution]
(i) Number of ways that 6 couples are seated together in a circle
6
5!: Group a couple as 1 unit. Arrange 6 units at a round table
6
2 : Within the unit, the couple has 2 ways to permute.
So for 6 units, we have
6
2 2 2 2 2 2 2
(ii) No. of ways that 12 people are seated in three distinct rows of four seats without
restriction 12!
No. of ways that 12 people are seated in a row with 0 female in one of the rows
3 6
1 4
C C 4! 8!
Hence required no. of ways
3 6
1 4
12! C C 4! 8!
3
1
C : Choose the row with 0 females
6
4
C : Choose 4 males to sit in that row
4! : Permutate the 4 chosen males in that row
8! : Permutate the remaining 8 people
4 (a)^ Find the number of 4-letter code-words that can be formed from the thirteen letters
of the word MEDITERRANEAN such that the first letter is E and the last letter is
R. [3]
[2009/IJC/Prelim/II/5]
(b) From the letters of the word DISTRIBUTION, find
(i) the number of 4-letter code-words that can be formed if the code-word contains
exactly three ‘I’s. [2]
(ii) the number of code-words that can be formed using all the letters such that all
the three ‘I’s are separated. [2]
[2010/JJC/Prelim/II/5]
[Solution]
(a) MEDITERRANEAN
E _ _ R (left M, D, I, T, EE, R, AA, NN)
Case 1: The letters in the centre are different
No. of ways =
8
2
C 2! 56 or
8 7
1 1
C C 8 7
Case 2: Both letters are the same
No. of ways =
3
1
C 3 [Choose from E, A or N]
Total no. of ways = 56 3 59
(b) (i) The different letters (other than I) left are D S T R B U O N
No. of different code words
8
1
C
(ii) No. of ways to permute the remaining 9 letters =
No. of ways to slot in the 3Is =
10
3
C
Thus number of code words that can be formed =
10
3
C
2 ‘Ts’ are
the same
A committee consisting of six people is to be selected from five women and six men.
Three of the women are sisters. Find the number of ways the committee can be
formed if
(i) the chosen committee will contain exactly two men, [1]
(ii) at least one of the sisters are included. [2]
The chosen committee consists of two particular sisters together with 3 other men and
one other woman. They are seated at a round table meant for six persons. Find the
number of possible arrangements if
(iii) one of the men is to be seated between the two sisters, [2]
(iv) the two sisters are sitting directly opposite each other. [2]
[2012/NYJC/Prelim/II/6]
5 [Solution]
(i) No. of ways =
6 5
2 4
C C 75
(ii) Number of ways if at least one of the sisters are included
= number of ways without restriction – number of ways if none of the sisters is
included
11 8
6 6
C C 434
Alternative solution [Consider cases]
Number of ways if at least one of the sisters are included
= no. of ways with 1 sister only + no. of ways with 2 sisters + no. of ways with all 3
sisters
3 8 3 8 3 8
1 5 2 4 3 3
C C C C C C
(iii) Select a man to be between the 2 sisters and group the 3 of them as one unit.
2! ways to arrange the 2 sisters. Then arrange the 4 units round a table.
Number of arrangements =
3
1
( C 2! ) ( 4 1)! 36
(iv) First seat the sisters directly opposite each other. Then arrange the 4 other
people.
Number of arrangements = 4! 24
7 A family of five wishes to go for lunch at their favourite restaurant. They hopped into
their family car which has two front seats and enough space for three persons behind.
Find the number of seating arrangements in the car if only the father can drive. [1]
At the restaurant, the family was shown to a round table with five seats. Find the number
of ways the family members could seat themselves if the youngest child insisted on
sitting with her mother. [2]
If the family was asked to sit at a round table with six seats which were numbered from
1 to 6 instead, find how many ways the family members could seat themselves if the
youngest child still insisted on sitting with her mother. [3]
[2009/ACJC/Prelim/II/5]
[Solution]
The number of seating arrangements = 4! = 24
Consider the youngest child and her mother as one unit and arrange it together with the
other 3 members of the family (4 distinct units) in a circle. The child and her mother
can exchange seats.
Number of ways = (4−1)! 2! = 12
The above 4 units, together with the empty seat, can be arranged in a circle. The child
and her mother can exchange seats. As the seats are numbered, each rotation around
the table gives a mutually exclusive case of arrangement.
Number of ways = (5−1)! 2! 6 = 288
8 Joshua tries to recall the 6-digit pin number for his ATM card.
(Assuming that the number 000 000 is even and valid.)
How many possible numbers can there be if he remembers that
(i) the number starts and ends with the digit 5? [1]
(ii) the number is odd and the digits do not repeat? [2]
(iii) the digits do not repeat and there are exactly 3 odd digits? [2]
[2011/PJC/Prelim/II/10]
[Solution]
(i) Number of way required = 10
4 = 10 000
(ii) Number of ways for last digit =
5
1
C = 5
Number of ways required =
5 9
1 5
C P 75 600 or 9 8 7 6 5 5
(iii) Number of ways for 3 odd digits = 3
5
C (^) = 10
Number of ways for 3 even digits = 3
5
C = 10
Number of ways required = 10 10 6! = 72 000
(a)(iii) Number of codes in which at least two letters L are together
= total no. of codes w/o restriction no. of codes where all the Ls
are separated
8
3
C
Alternative solution
Case 1: 2 letters L together
Number of ways =
8
2
C
Case 2: All letters L together
Number of ways =
Thus, required number of ways = 141 120 + 20 160 = 161 280
P(remaining seven cards left consist of all different letters)
= P(the 3 cards drawn are L, L, P)
: Number of cases [Case 1: L, L, P Case 2: L, P, L Case 3: P, L, L]
Alternative solution
P(remaining seven cards left consist of all different letters)
3 2
2 1
10
3
C C
C
Slotting in method for LL and L to
ensure that they are separated
10 (a) In how many ways can 7 identical packs of chocolates be distributed among 20
children, if no child can get more than one pack? [1]
(b) In how many ways can 4 different gifts be distributed among 20 children if each
child can get any number of gifts? [1]
(c) 9 people go to a restaurant with the layout
given by the diagram on the right. In how
many ways can the 9 people be seated on the
chairs marked A, B, C, …, H, I? [1]
Find the number of ways in which the 9 people can be seated if
(i) 3 particular people do not want to be seated on any of the 4 chairs A, D, E and
I, in front of windows. [2]
(ii) 2 particular people do not want to be seated next to each other on the same side
of the table. [3]
[modified 2009/SRJC/Prelim/II/11]
[Solution]
(a) Since the chocolate packs are identical, the order does not matter (cannot
distinguish which child gets which pack). Choose a group of 7 children from the
20 children to receive a pack each.
No. of ways =
(b) Each of the 4 gifts can be given to any of the 20 children (20×20×20×20). Note
that the gifts are different, so the order matters.
No. of ways =
4
(c) Number of ways 9 people can be seated = 9! = 362880
(i) Number of ways the 3 particular people can be seated
5
3
P 60
Number of ways the 6 other people can be seated = 6!
Total ways = 60 × 6! = 43200
Alternative solution
No. of ways the 6 other people can take up the window seats
6
4
C 4! 360
Total ways = 360 ×5! = 43200
(ii) No. of ways the two person are seated together on the side with seats labelled
A to D = (^) 3 2! 7!
3 ways: A & B, B & C, C & D
12 A test consists of five Pure Mathematics questions, A, B, C, D and E, and six Statistics
questions, F, G, H, I, J and K.
(i) The examiner plans to arrange all eleven questions in a random order, regardless
of topic. Find the number of ways to arrange all eleven questions such that
(a) the last question is a Pure Mathematics question, [2]
(b) a Pure Mathematics question must be separated from another with exactly one
Statistics question. [2]
(ii) Later, the examiner decides that the questions should be arranged in two sections,
Pure Mathematics followed by Statistics. Find the number of ways to arrange all
eleven questions such that
(a) question A is followed by question F, [2]
(b) questions B and K are separated by more than seven questions. [3]
[2010/DHS/Prelim/II/8]
[Solution]
(i) (a)
5
1
Number of ways C10! 18144 000
(i) (b)
S S S S S S
Number of ways (^) 6! (^) 5! (^) 3 259 200
Alternative solution
Number of ways =
6
4
C 4! 5! 3! 259 200
PSPSPSPSP S S
12 (ii)(a)
A F
Number of ways (^) 4! 5! 2880
(ii)(b) Case 1:
8 Questions (4P and 4S) between B and K
B K
Case 2:
8 Questions (3P and 5S) between B and K
B K
Case 3:
9 Questions between B and K
B K
Pure Mathematics Question Statistics Question
No. of ways = 4! 4! 3! 5! 5! 4!
4!
5!
4! 5!
14 Mr and Mrs Tan have a son named David. They invite 6 guests to their home for dinner.
Before the dinner, they plan to take a group photo together with the guests. They are to
stand in a straight row for the photo-taking.
Find the number of arrangements for the photo-taking if
(i) all the 9 diners can stand in any position,
(ii) the Tan family members must be separated.
The Tan family decides to seat the diners at two round tables – a five-seater and a smaller
four-seater. Find the number of arrangements this can be done if
(iii) anyone can be seated at either table,
(iv) David must be seated between his parents at either table.
[2011/SAJC/Prelim/II/6]
[Solution]
(i) No. of ways= 9! 362 880
(ii) Arrange the 6 non-Tan diners first. No of ways to do that = 6!
We can slot the 3 Tans in between the diners in
7
3
P or
7
3
C 3!ways.
Total no of ways= 6!
7 P 3 = 151 200
(iii) No. of ways to choose diners for the four-seater =
9
4
C = 126
No. of ways to seat the diners= (4−1)!= 3!
No. of ways to choose diners for five-seater =
5
5
C =
No. of ways to seat the diners= (5−1)!=4!
Total no. of ways =
9 5
4 5
C 3! C 4!= 18144
(iv) Case 1: David is seated at four-seater
No. of ways to sit David at four-seater = (^)
6
1
C 2 1! 2! 12
No. of ways to seat the rest at five-seater = (5−1)! = 4! =
Total no. of ways (for Case 1) = 12 × 24 = 288
Case 2: David is seated at five-seater
No. of ways to seat David at five-seater =
6
2
C 3 1! 2! 60
No. of ways to seat the rest at four-seater = (4−1)! =3! = 6
Total no. of ways (For Case 2) = 6 × 60 = 360
Total no. of ways= 288+360= 648
15 Mary Lim has 7 cousins. In how many ways can she invite some or all of them to her
birthday party? [2]
At her birthday party, Mary sets up a round table of 8 seats with a different welcome
gift at each seat. If all her cousins turn up for the party and given that 4 of them are from
the Lee family, 2 are from the Yeo family and 1 is from the Tan family, find the number
of ways they can be seated with Mary for a meal at the table if family members of the
same surname are seated together but members of the Lee and Yeo families are not
adjacent to each other. [3]
After the meal, Mary and her cousins start to play a game. The game requires a
formation of 2 facilitators and 2 teams of 3 members each. Find the number of possible
formations if not all the members in each team have the same surname. [3]
[2010/AJC/Prelim/II/6]
[Solution]
For each of the cousin, they are either invited or not (2 outcomes) and she must have
invited at least one.
No. of ways to invite her guests =
7
2 1 = 127
Mary Lim and Tan must each be seated between the Lee and Yeo families.
Note that the different welcome gifts at each seat is equivalent to having the seats
labelled.
No. of ways to arrange a round table
No. of selections of 2 facilitators and 2 teams (without any restrictions)
8 C 2 (
6 C 3
3 C 3 ) 2! = 280
No. of selections of 2 facilitators and 2 teams with Lee family forming 1 team
4 C 3
5 C 2
3 C 3 = 40
No. of possible formations = 280 – 40 = 240
permutating the Lee and Yeo famililies
17 A group of 12 students consisting of 6 girls and 6 boys are to be seated at tables A, B
and C, each with 4 chairs numbered 1, 2, 3 and 4 as shown below.
Find the number of ways this can be done if
(i) there is no restriction, [1]
(ii) there are 2 girls and 2 boys at each table, [3]
(iii) there are 2 girls and 2 boys at each table and students of the same gender are seated
opposite each other. [2]
[2008/RJC/Prelim/II/6]
[Solution]
(i) The seats can be seen as a “row” of seats A1, A2, A3, A4, B1, …, C
Number of ways = 12! = 479 001 600
Alternative solution
Number of ways =
12 8 4
4 4 4
C 4! C 4! C 4!
3 12 8 4
4 4 4
C C C 4!
(ii) Number of ways
6 6 4 4 2 2
2 2 2 2 2 2
C C 4! C C 4! C C 4!
^
2 3 6 4 2
2 2 2
C C C 4!
(iii) Number of ways
^ ^ ^ ^ ^
2 2 2 6 4 2
2 2 2
C 2 2! 2! C 2 2! 2! C 2 2! 2!
^
2 3 6 4 2
2 2 2
C C C 2 2! 2!
1
4
3
2 A
1
4
3
2 B
1
4
3
2 C
The pair of girls could be
seated at 1,3 or 2,4.
18 An electronic game called ‘Wishful Thinking’ is played with 5 boxes arranged in a row.
When a button is pressed, each of the five boxes displays a picture of a fruit – either an
apple, orange or pear. A possible result of the game is shown below.
Events ‘Success’ and ‘Windfall’ are defined as follows.
‘Success’ : Exactly 3 boxes display the same fruit.
‘Windfall’ : All 5 boxes display the same fruit.
Find
(i) the total number of possible results, [1]
(ii) the number of ways of obtaining ‘Windfall’, [1]
(iii) the total number of ways of not obtaining ‘Success’. [3]
Ten electronic game machines with distinct serial numbers are sent to 3 different game
centres. In how many ways can the game machines be distributed if each centre must
have at least 3 machines? [3]
[2010/IJC/Prelim/II/9]
[Solution]
(i) Total number of possible results =
(ii) Number of ways of obtaining ‘Windfall’ = 3
(iii) No. of ways to obtain a success =
5
3
C 3 2 2 = 120
Total number of ways of not getting a success
Required no. of ways =
10 7 4
3 3 4
C C C 3 12600
3 : 3 ways
Game Centre A, B, C
5 3 243
Box 1 Box 2 Box 3 Box 4^ Box 5
Choose 3 slots
to have the
same fruit.
The remaining 2 slots each
can have either of the 2
remaining type of fruits.
