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In this lesson, we will learn how to use the properties of permutations to simplify expressions and solve equations.

Q1:

Evaluate 9 5 π .

Q2:

Evaluate 6 3 7 5 π + π .

Q3:

Evaluate 5 5 2 2 π Γ π .

Q4:

Evaluate 7 4 5 2 π π .

Q5:

Find the value of π given that π β 1 0 2 π = 3 .

Q6:

If 4 9 π + 3 4 9 π + 2 π = 3 4 Γ π , find the value of π β 6 .

Q7:

If π π π = 7 2 0 and π = 1 2 0 , find the value of 9 π β 7 π .

Q8:

Find the solution set of the equation 2 4 0 π = π π₯ + 2 2 π₯ + 4 4 .

Q9:

Find the value of π such that π 3 π = 3 2 7 3 6 .

Q10:

If π + 1 π β 1 = 7 2 , find π 5 π 6 π 7 π + π + π .

Q11:

Find the smallest value of π satisfying the inequality π 2 5 π 2 4 π > π .

Q12:

Find π such that οͺ ο¦ ο ο° ο§ ο« ο§ ο§ ο ο° ο§ ο« π = π .

Q13:

If π 1 5 π β 1 1 4 π = 2 3 οΉ π ο , find π .

Q14:

If π₯ π π = 1 2 0 and π = 2 4 , find π₯ + 2 π β 2 π .

Q15:

If 1 2 π₯ π = 1 3 2 0 , find 7 π₯ β 5 π₯ π .

Q16:

Find the value of π such that ο§ ο ο π = 4 , 0 8 0 .

Q17:

If ο ο± ο ο© π = 3 5 , 9 0 4 and π + π = 6 0 , find π and π .

Q18:

Evaluate ( π + 2 ) ! given that 8 π π = 3 3 6 .

Q19:

If οΌ 3 4 π ο ! = 7 2 0 , find the value of π 5 π .

Q20:

Find the value of π₯ satisfying the equation 1 0 7 π₯ 1 0 7 π₯ β 1 π β 5 π₯ π = 0 .

Q21:

Given that π 4 π β 1 3 π = 8 Γ π , determine the value of π .

Q22:

If π₯ + π¦ 4 π = 3 6 0 and ( 2 π₯ + π¦ ) ! = 5 0 4 0 , find π¦ π₯ π .

Q23:

Evaluate 5 2 π Γ 2 ! .

Q24:

If the middle factor in the expansion of π 2 3 π is 69, find the value of π .

Q25:

Find the value of the permutation for π = 6 and π = 2 .

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