SOLVING EQUATIONS INVOLVING PERMUTATIONS

Problem 1 :

Solve for n :

Solution :

n² - n - 42 = 0

Factoring this quadratic equation, we get

(n - 7)(n + 6) = 0

Equating each factor to 0, we get

n - 7 = 0 and n + 6 = 0

n = 7 and n = -6

Here -6 is not admissible. So, the answer is 7.

Problem 2 :

Solution :

Removing factorial on both sides of the equal sign, we get

5 = 7 - r

r = 7 - 5

r = 2

Problem 3 :

Solution :

Cancelling (n-2)! on both numerator and denominator, we get

n(n-1) = 30

n² - n - 30 = 0

(n - 6)(n + 5) = 0

n = 6 and n = -5

So, the value of n is 6.

Problem 4 :

Solution :

Canceling (n - 3)! on both numerator and denominator, we get

(n - 1)(n - 2) = 12

n² - 3n + 2 = 12

n² - 3n + 2 - 12 = 0

n² - 3n - 10 = 0

(n - 5) (n + 2) = 0

n = 5 and n = -2

So, the value of n is 5.

Problem 5 :

Solution :

(7 - r)(6 - r) = 6

42 - 7r - 6r + r² = 6

r² -13r + 42 - 6 = 0

r² -13r + 36 = 0

(r - 9)(r - 4) = 0

r = 9 and r = 4

So, the value of r is 4.

Problem 6 :

Solve for n

Solution :

(6 - n)(5 - n) = 6

30 - 6n - 5n + n² = 6

n²- 11n + 30 - 6 = 0

n²- 11n + 24 = 0

(n - 3)(n - 8) = 0.

n = 3 and n = 8

So, the value of n is 3.

Problem 7 :

Solve for n

Solution :

(n-3)(n-4) = 20

n² - 3n - 4n + 12 = 20

n² - 7n + 12 - 20 = 0

n² - 7n - 8 = 0

(n - 8) (n + 1) = 0

n = 8 and n = -1

So, the value of n is 8.

Problem 8 :

Solution :

Problem 9 :

Solution :

Problem 10 :

Solution :

156 - 25r + r² = 12

r²- 25r + 156 - 12 = 0

r²- 25r + 144 = 0

(r - 16)(r - 9) = 0

r = 16 and r = 9

So, the value of r is 9.

Problem 11 :

Solution :

(n - 2)(n - 3) = 12

n² - 5n + 6 = 12

n² - 5n + 6 - 12 = 0

n² - 5n - 6 = 0

(n - 6)(n + 1) = 0

n = 6 and n = -1

So, the value of n is 6.