Short Question 9 & 10

Posted on May 18, 2020 by Myhometuition

Question 9 (2 marks):

\begin{array}{l} (a) Given {}^{6}C_{n} > 1, list out all the \\ possible values of n . \\ (b) Given {}^{y}C_{m} = {}^{y}C_{n}, express y in \\ terms of m and n . \end{array}

Solution:
(a)
n = 1, 2, 3, 4, 5

(b)
y = m + n

Question 10 (4 marks):
Danya has a home decorations shop. One day, Danya received 14 sets of cups from a supplier. Each set contained 6 pieces of cups of different colours.

(a) Danya chooses 3 sets of cups at random to be checked.
Find the number of different ways that Danya uses to choose those sets of cups.

(b) Danya takes a set of cups to display by arranging it in a row.
Find the number of different ways the cups can be arranged such that the blue cup is not displaced next to the red cup.

Solution:
(a)
Number of different ways 3 sets of cups at random to be checked
= ¹⁴C₃=364

(b)

Number of ways (Blue cup and red cup are next to each other)
= 5! × 2!
= 240

Number of different ways the cups can be arranged such that the blue cup is not displaced next to the red cup
= 6! – 240
= 720 – 240
= 480

Short Question 7 & 8

Posted on May 17, 2020 by Myhometuition

Question 7:
Six members of a committee of a school are to be selected from 6 male teachers, 4 female teachers and a male principal. Find the number of different committees that can be formed if
(a) the principal is the chairman of the committee,
(b) there are exactly 2 females in the committee,
(c) there are not more than 4 males in the committee.

Solution:
(a)
If the principal is the chairman of the committee, the remaining number of committee is 5 members.
Hence, the number of different committees that can be formed from the remaining 6 male teachers and 4 female teachers

\begin{array}{l} =^{10} C_{5} \\ = 252 \end{array}

(b)

\begin{array}{l} Exactly 2 females in the committee \\ =^{4} C_{2} \times^{7} C_{4} \\ = 210 \end{array}

(c)

\begin{array}{l} There are not more than 4 males in the committee \\ = 4 males 2 females + 3 males 3 females + 2 males 4 females \\ =^{7} C_{4} \times^{4} C_{2} +^{7} C_{3} \times^{4} C_{3} +^{7} C_{2} \times^{4} C_{4} \\ = 210 + 140 + 21 \\ = 371 \end{array}

Question 8:

The diagram below shows five cards of different letters.

R E A C T

(a) Calculate the number of arrangements, in a row, of all the cards.

(b) Calculate the number of these arrangements in which the letters E and A are side by side.

Solution:
(a) Number of arrangements = 5! = 120

(b)
Step 1
If the letters ‘E ’ and ‘A’ have to be placed side by side, they will be considered as one item.
Together with the letters ‘R ’, ‘C ’ and ‘T ’, there are altogether 4 items.

EA R C T

Number of arrangements = 4!

Step 2

The letters ‘E ’ and ‘A ’ can also be arranged among themselves in their group.
Number of arrangements = 2!

Hence, number of arrangements of all the letters of the word ‘REACT’ in which the letters E and A have to be side by side
= 4! × 2!
= 24 × 2
= 48

6.1 Permutation Part 1

Posted on April 23, 2020 by user

(A) rs Multiplication Principle/ Rule

1. If an operation can be carried out in r ways and another operation can be carried out in s ways, then the number of ways to carry out both the operations consecutively is r × s, i.e. rs.

2. The rs multiplication principle can be expanded to three or more operations. If the numbers of ways for the occurrence of events A, B and C are r, s and p respectively, the number of ways for the occurrence of all the three events consecutively is r × s × p, i.e. rsp.

Example 1:
There are 3 different roads to travel from town P to town Q and 4 different roads to travel from town Q to town R. Calculate the number of ways a person can travel from town P to town R via town Q.

Solution:

3 × 4 = 12

(B) Permutations

Example 2:
Calculate each of the following.
(a) 7!
(b) 4!6!
(c) 0!5!

\begin{array}{l} (d) \frac{7!}{5!} \\ (e) \frac{8!}{4!} \\ (f) \frac{n!}{(n - 2)!} \\ (g) \frac{n! 0!}{(n - 1)!} \\ (h) \frac{3! (n + 1)!}{2! n!} \end{array}

Solution:

(a) 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

(b) 4!6! = (4 × 3 × 2 × 1)( 6 × 5 × 4 × 3 × 2 × 1) = 17280

(c) 0!5! = (1)( 5 × 4 × 3 × 2 × 1) = 120

\begin{array}{l} (d) \frac{7!}{5!} = \frac{7 \times 6 \times 5!}{5!} = 7 \times 6 = 42 \\ (e) \frac{8!}{4!} = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4!} = 8 \times 7 \times 6 \times 5 = 1680 \\ (f) \frac{n!}{(n - 2)!} = \frac{n (n - 1) (n - 2)}{(n - 2)} = n (n - 1) \\ (g) \frac{n! 0!}{(n - 1)!} = \frac{n (n - 1) (1)}{(n - 1)} = n \\ (h) \frac{3! (n + 1)!}{2! n!} = \frac{3 \times 2! (n + 1) (n) (n - 1)}{2! n (n - 1)} = 3 (n + 1) \end{array}

Calculator Computation:

Short Question 1 – 3

Posted on April 23, 2020 by user

Question 1:
A group of 4 men and 3 ladies are to be seated in a row for a photographing session. If the men and ladies want to be seated alternately (man-lady-man-lady...), calculate the number of different arrangements.

Solution:
The arrangements of 4 men and 3 ladies to be seated alternately are as follow:

M L M L M L M

The number of ways to arrange the seat for 4 men = 4!

The number of ways to arrange the seat for 3 ladies = 3!

Total number of different arrangements for men and ladies = 4! × 3! = 144

Question 2:
Ahmad has 6 durians, 5 watermelons and 2 papayas. If he wants to arrange these fruits in a row and the fruits of the same kind have to be grouped together, calculate the number of different arrangements. The sizes of the fruits are different.

Solution:
The number of ways to arrange 3 groups of fruits that are same kind = 3!

DDDDDD WWWWW PP

The number of ways to arrange 6 durians = 6!
The number of ways to arrange 5 watermelons = 5!
The number of ways to arrange 2 papayas = 2!

Therefore, the number of ways to arrange the fruits with same kind of fruits was grouped together
= 3! × 6! × 5! × 2!
= 1036800

Question 3:
Calculate the number of arrangements, without repetitions, of the letters from the word `SOMETHING' with the condition that they must begin with a vowel.

Solution:

\begin{array}{l} Arrangement of letters begin with vowel \\ = O, M and I \\ =^{3} P_{1} \\ Arrangement for the rest of the letters \\ = 7! \\ ∴ Number of arrangements \\ =^{3} P_{1} \times 7! \\ = 15120 \end{array}

6.3 Combinations

Posted on April 23, 2020 by user

6.3 Combinations

(1) The number of combinations of r objects chosen from n different objects is given by :

^{n} C_{r} = \frac{n!}{r! (n - r)!}

(2) A combination of r objects chosen from n different objects is a selection of a set of r objects chosen from n objects. The order of the objects in the chosen set is not taken into consideration.

\begin{array}{l} N o t e : \\ (i)^{n} C_{0} = 1 \\ (i i)^{n} C_{n} = 1 \\ (i i i)^{n} C_{r} =^{n} C_{n - r} \end{array}

Example 1:

\begin{array}{l} {Calculate the value of}^{7} C_{2} \\ ^{7} C_{2} = \frac{7!}{(7 - 2)! \times 2!} \\ = \frac{7!}{5! \times 2!} \\ = \frac{7 \times 6 \times 5!}{5! \times 2!} \\ = \frac{7 \times 6}{2 \times 1} \\ = 21 \end{array}

Calculator Computation:

Example 2:

There are 6 marbles, each with different colour, which are to be divided equally between 2 children. Find the number of different ways the division of the marbles can be done.

Solution:

Number of ways giving 3 marbles to the first child =

^{6} C_{3}

Number of ways giving the remaining 3 marbles =

^{3} C_{3}

So, the number of different ways the division of the marbles

\begin{array}{l} =^{6} C_{3} \times^{3} C_{3} \\ = 20 \times 1 \\ = 20 \end{array}

6.1 Permutations Part 2

Posted on April 23, 2020 by user

(C) Permutation of n Different Objects, Taken r at a Time

1. The number of permutations of n different objects, taken r at a time is given by:

2. A permutation of n different objects, taken r at a time, is an arrangement of a set of r objects chosen from n objects. The order of the objects in the chosen set is taken into consideration.

3. The number of permutations of n different objects, taken all at a time, is:

Example 1:

Evaluate each of the following.

\begin{array}{l} {(a)}^{5} P_{2} \\ {(b)}^{7} P_{3} \\ {(c)}^{9} P_{4} \end{array}

Solution:

\begin{array}{l} {(a)}^{5} P_{2} = \frac{5!}{(5 - 2)!} = \frac{5!}{3!} \\ = \frac{5 \times 4 \times 3!}{3!} = 5 \times 4 = 20 \\ {(b)}^{7} P_{3} = \frac{7!}{(7 - 3)!} = \frac{7!}{4!} \\ = \frac{7 \times 6 \times 5 \times 4!}{4!} = 7 \times 6 \times 5 = 210 \\ {(c)}^{9} P_{4} = \frac{9!}{(9 - 4)!} = \frac{9!}{5!} \\ = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5!} = 9 \times 8 \times 7 \times 6 = 3024 \end{array}

Calculator Computation:

Short Question 5 – 8

Posted on April 23, 2020 by user

Question 4:
A committee that consists of 6 members is to be selected from 5 teachers and 4 students. Find the number of different committees that can be formed if
(a) there is no restriction,
(b) the number of teachers must exceed the number of students.

Solution:
(a)
Total number of committees = 5 + 4 = 9
6 members to be selected from 9 committees with no restriction

^{= 9} C_{6} = 84

(b)

\begin{array}{l} If the number of teachers must exceed the \\ number of students, the combination \\ = 4 teachers 2 students + 5 teachers 1 student \\ =^{5} C_{4} \times^{4} C_{2} +^{5} C_{5} \times^{4} C_{1} \\ = 30 + 4 \\ = 34 \end{array}

Question 5:
A school prefect committee that consists of 6 persons is to be chosen from 6 Malays, 5 Chinese and 4 Indians. Calculate the number of different committees that can be formed if the number of Malays, Chinese and Indians must be equal.

Solution:
Number of different committees that can be formed for 2 Malays, 2 Chinese and 2 Indians

\begin{array}{l} =^{6} C_{2} \times^{5} C_{2} \times^{4} C_{2} \\ = 900 \end{array}

Question 6:
There are 10 different flavour candies in a plastic bag.
Find
(a) the number of ways 3 candies can be chosen from the plastic bag.
(b) the number of ways at least 8 candies can be chosen from the plastic bag.

Solution:
(a)
Number of ways choosing 3 candies out of 10 candies

\begin{array}{l} =^{10} C_{3} \\ = 120 \end{array}

(b)
Number of ways choosing 8 candies =

=^{10} C_{8}

Number of ways choosing 9 candies =

^{10} C_{9}

Number of ways choosing 10 candies =

^{10} C_{10}

Hence, number of ways of choosing at least 8 candies

\begin{array}{l} =^{10} C_{8} +^{10} C_{9} +^{10} C_{10} \\ = 56 \end{array}