Question 10:
(a) Determine the number of different ways to arrange 7 letters from the word ‘SEMPURNA’ in a row without repetition. [1 mark]
(b) A four-digit number is to be formed by using the digits 1, 2, 3, 4, 5 and 6, such that the repetition of digits 1, 2, 3, 4 and 5 is not allowed.
Find the number of different ways the four-digit numbers can be formed.
[3 marks]
Answer:
(a) $$ \begin{aligned} \text { Number of ways } & ={ }^8 P_7 \\ & =40320 \end{aligned} $$
(b) $$ \begin{aligned} & \text { Case 1: No repetition }\\ &{ }^6 P_4=360 \end{aligned} $$
Case 2: Repeat digit 6 two times (6, 6)
$$ { }^4 C_2 \times{ }^5 P_2=120 $$
Case 3: Repeat digit 6 three times (6, 6, 6)
$$ { }^4 C_3 \times{ }^5 P_1=20 $$
$$ \begin{aligned} & \text { Case 4: } \text { All digit } 6 \text { (6, 6, 6, 6) }\\ &{ }^4 C_4=1 \end{aligned} $$
$$ \begin{aligned} \text { Total number of ways } & =360+120+20+1 \\ & =501 \end{aligned} $$
(a) Determine the number of different ways to arrange 7 letters from the word ‘SEMPURNA’ in a row without repetition. [1 mark]
(b) A four-digit number is to be formed by using the digits 1, 2, 3, 4, 5 and 6, such that the repetition of digits 1, 2, 3, 4 and 5 is not allowed.
Find the number of different ways the four-digit numbers can be formed.
[3 marks]
Answer:
(a) $$ \begin{aligned} \text { Number of ways } & ={ }^8 P_7 \\ & =40320 \end{aligned} $$
(b) $$ \begin{aligned} & \text { Case 1: No repetition }\\ &{ }^6 P_4=360 \end{aligned} $$
Case 2: Repeat digit 6 two times (6, 6)
$$ { }^4 C_2 \times{ }^5 P_2=120 $$
Case 3: Repeat digit 6 three times (6, 6, 6)
$$ { }^4 C_3 \times{ }^5 P_1=20 $$
$$ \begin{aligned} & \text { Case 4: } \text { All digit } 6 \text { (6, 6, 6, 6) }\\ &{ }^4 C_4=1 \end{aligned} $$
$$ \begin{aligned} \text { Total number of ways } & =360+120+20+1 \\ & =501 \end{aligned} $$
Question 11:
There are 25 students present at the general meeting of STEM Society.
(a) Find the number of different ways to choose 4 students to lead the committee. [1 mark]
(b) After the leaders are chosen, the remaining number of boys to the number of girls is in the ratio 4 : 3. A boy called Zam and a girl called Min are among them.
Find the number of different ways 5 committee members can be chosen, such that the number of boys is more than the number of girls and must include either Zam or Min.
[3 marks]
Answer:
(a) $$ \begin{aligned} \text { Number of ways } & ={ }^{25} C_4 \\ & =12650 \end{aligned} $$
(b) $$ \begin{aligned} &\text { Remaining number of students }\\ &\begin{aligned} & =25-4 \\ & =21 \end{aligned} \end{aligned} $$
$$ B>G \rightarrow 3 B+2 G \text { or } 4 B+1 G $$
3B + 2G
$$ \text { Case } 1 \text { :Include Zam only → Zam }+2 B+2 G $$
$$ \text { Case } 2 \text { :Include Min only } \text { → } \text { Min }+3 B+1 G $$
$$ \text { Case } 3 \text { :Include both → Zam }+ \text { Min }+2 B+1 G $$
$$ \begin{aligned} \text { Number of ways }= & \left({ }^1 C_1 \times{ }^{11} C_2 \times{ }^9 C_2\right)+ \\ & \left({ }^1 C_1 \times{ }^{12} C_3 \times{ }^8 C_1\right)+ \\ & \left({ }^1 C_1 \times{ }^1 C_1 \times{ }^{11} C_2 \times{ }^8 C_1\right) \\ = & 1980+1760+440 \\ = & 4180 \end{aligned} $$
4B + 1G
$$ \text { Case } 1 \text { :Include Zam only → Zam }+3 B+1 G $$
$$ \text { Case } 2 \text { :Include Min only ⟶ Min + } 4 B $$
$$ \text { Case } 3 \text { :Include both → Zam + Min + 3B } $$
$$ \begin{aligned} \text { Number of ways }= & \left({ }^1 C_1 \times{ }^{11} C_3 \times{ }^9 C_1\right)+ \\ & \left({ }^1 C_1 \times{ }^{12} C_4\right)+ \\ & \left({ }^1 C_1 \times{ }^1 C_1 \times{ }^{11} C_3\right) \\ = & 1485+495+165 \\ = & 2145 \end{aligned} $$
$$ \begin{aligned} \text { Total number of ways } & =4180+2145 \\ & =6325 \end{aligned} $$
There are 25 students present at the general meeting of STEM Society.
(a) Find the number of different ways to choose 4 students to lead the committee. [1 mark]
(b) After the leaders are chosen, the remaining number of boys to the number of girls is in the ratio 4 : 3. A boy called Zam and a girl called Min are among them.
Find the number of different ways 5 committee members can be chosen, such that the number of boys is more than the number of girls and must include either Zam or Min.
[3 marks]
Answer:
(a) $$ \begin{aligned} \text { Number of ways } & ={ }^{25} C_4 \\ & =12650 \end{aligned} $$
(b) $$ \begin{aligned} &\text { Remaining number of students }\\ &\begin{aligned} & =25-4 \\ & =21 \end{aligned} \end{aligned} $$
$$ B>G \rightarrow 3 B+2 G \text { or } 4 B+1 G $$
3B + 2G
$$ \text { Case } 1 \text { :Include Zam only → Zam }+2 B+2 G $$
$$ \text { Case } 2 \text { :Include Min only } \text { → } \text { Min }+3 B+1 G $$
$$ \text { Case } 3 \text { :Include both → Zam }+ \text { Min }+2 B+1 G $$
$$ \begin{aligned} \text { Number of ways }= & \left({ }^1 C_1 \times{ }^{11} C_2 \times{ }^9 C_2\right)+ \\ & \left({ }^1 C_1 \times{ }^{12} C_3 \times{ }^8 C_1\right)+ \\ & \left({ }^1 C_1 \times{ }^1 C_1 \times{ }^{11} C_2 \times{ }^8 C_1\right) \\ = & 1980+1760+440 \\ = & 4180 \end{aligned} $$
4B + 1G
$$ \text { Case } 1 \text { :Include Zam only → Zam }+3 B+1 G $$
$$ \text { Case } 2 \text { :Include Min only ⟶ Min + } 4 B $$
$$ \text { Case } 3 \text { :Include both → Zam + Min + 3B } $$
$$ \begin{aligned} \text { Number of ways }= & \left({ }^1 C_1 \times{ }^{11} C_3 \times{ }^9 C_1\right)+ \\ & \left({ }^1 C_1 \times{ }^{12} C_4\right)+ \\ & \left({ }^1 C_1 \times{ }^1 C_1 \times{ }^{11} C_3\right) \\ = & 1485+495+165 \\ = & 2145 \end{aligned} $$
$$ \begin{aligned} \text { Total number of ways } & =4180+2145 \\ & =6325 \end{aligned} $$