Question

If n(A) =3 and n(B)=5 then the number of possible one -one functions from A-> B is

Answer 1

Answer:

n(A)=3

n(B)=5

NUM of possible outcomes=n(A)*n(B)

=3*5

=15

Answer 2

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$n(A) =3 \: \: and \: \: n(B)=5$

So the required number of possible one -one functions

=$\frac{5! }{(5 - 3)! } = \frac{5!}{ \: 2! } = \frac{5 \times 4 \times 3 \times 2! }{2! } = 60$

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