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In this lesson, we will learn how to use rules of combinations, such as ratio, addition, and simplification rules.

Q1:

Determine the value of 2 1 7 2 1 6 πΆ πΆ without using a calculator.

Q2:

Determine the value of 2 3 8 2 3 6 πΆ πΆ without using a calculator.

Q3:

Given that π + 1 9 π₯ + 1 9 π + 1 9 π₯ + 1 8 πΆ βΆ πΆ = 2 βΆ 1 , determine π .

Q4:

Given that , determine and .

Q5:

Find π₯ and π¦ , such that π₯ π¦ π₯ π¦ + 1 πΆ = 3 8 Γ πΆ and π₯ π¦ π₯ π¦ + 6 πΆ = πΆ .

Q6:

Find the value of π 5 π + 2 πΆ , given that π π + 4 π π 4 π β 5 πΆ = πΆ 2 and π 1 2 πΆ = 4 5 5 .

Q7:

Find all of the possible values of π , given that 4 Γ πΆ = 3 Γ πΆ + 2 Γ πΆ π 5 π 6 π 4 .

Q8:

Find the possible values of π which satisfy the equation 2 1 π 2 1 1 5 πΆ = πΆ .

Q9:

Using the equation π π π π β 1 π π β 2 π + 2 π πΆ + 2 Γ πΆ + πΆ = πΆ , evaluate the expression 2 4 2 1 2 4 2 0 2 4 1 9 πΆ + 2 Γ πΆ + πΆ .

Q10:

Which of the following is equal to π 1 6 π 6 7 π 7 πΆ Γ πΆ πΆ ?

Q11:

If 1 3 2 π β 9 1 3 π + 1 πΆ = πΆ , determine π .

Q12:

If π 4 πΆ = 1 5 , determine 2 π π + 4 πΆ .

Q13:

Evaluate 7 2 8 6 πΆ πΆ .

Q14:

Evaluate 1 6 2 1 6 3 1 7 2 πΆ + πΆ πΆ .

Q15:

Find the value of π given that π 1 8 π 1 7 πΆ βΆ πΆ = 2 βΆ 1 .

Q16:

Determine π given that 2 4 ( πΆ ) = π π π π π .

Q17:

Evaluate 3 5 3 3 5 2 3 5 1 πΆ + 2 Γ πΆ + πΆ .

Q18:

Given that 2 9 3 π β 6 2 9 2 π πΆ = πΆ , determine all the possible values of π .

Q19:

If π 4 2 π π β 4 2 π 4 3 πΆ + πΆ = 2 πΆ , find π .

Q20:

Given that 7 0 7 1 7 2 7 3 7 4 7 5 7 6 7 7 π πΆ + πΆ + πΆ + πΆ + πΆ + πΆ + πΆ + πΆ = 2 , determine the value of π .

Q21:

If 2 7 π + 8 2 7 π + 9 πΆ βΆ πΆ = 6 βΆ 1 and π + 1 0 π β 1 1 π + 1 0 π β 1 0 πΆ + πΆ = 3 3 6 4 9 , find π and π .

Q22:

Find the value of π such that 1 5 2 π β 4 1 5 π + 3 πΆ = πΆ .

Q23:

By applying the relation π π π π β 1 π + 1 π πΆ + πΆ = πΆ , deduce the value of 5 9 2 5 9 3 πΆ + πΆ .

Q24:

Given that π π π π + 1 π π πΆ + πΆ πΆ = π + 1 π + 1 and π 2 1 π 2 2 π 2 1 πΆ + πΆ πΆ = 3 2 , find π .

Q25:

Write ο ο ο ο ο° ο§ πΆ + πΆ in the form οΊ ο» πΆ .

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