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Let A = {1, 2, 3, ..., 10} and f : A → A be defined as f(k) = is oddis even{k+1ifk is odd kifk is even. Then the number of possible functions g : A → A such that gof = f is ______. -

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Question

Let A = {1, 2, 3, ..., 10} and f : A `rightarrow` A be defined as

f(k) = `{{:(k + 1, if k  "is odd"),(     k, if k  "is even"):}`.

Then the number of possible functions g : A `rightarrow` A such that gof = f is ______.

Options

  • 55

  • 105

  • 5!

  • 10C5

MCQ
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Solution

Let A = {1, 2, 3, ..., 10} and f : A `rightarrow` A be defined as

f(k) = `{{:(k + 1, if k  "is odd"),(     k, if k  "is even"):}`.

Then the number of possible functions g : A `rightarrow` A such that gof = f is `underlinebb(10^5)`.

Explanation:

Putting value of K from 1 to 10, we get

f(1) = f(2) = 2

f(3) = f(4) = 4

f(5) = f(6) = 6

f(7) = f(8) = 8

f(9) = f(10) = 10

Since, g(f(x)) = f(x)

∴ gof(1) = f(1) `\implies` g(2) = f(1) = 2

gof(2) = f(2) `\implies` g(2) = f(2) = 2

gof(3) = f(3) `\implies` g(4) = f(3) = 4

∴ The image of 2, 4, 6, 8, 10 in function g(x) should be 2, 4, 6, 8, 10 respectively. Therefore, image of each of remaining elements can be any of 10 elements.

Hence, number of possible g(x) is 105.

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