Answer
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Hint:
Find the number of elements in A and B i.e. n(A) and n(B). Thus, no. of one-one function from A into B can be find by using \[{}^{n(B)}{P_{n(A)}}\]
Complete step by step solution:
Given,
A = {1, 3, 5, 7}
and, B = {1, 2, 3, 4, 5, 6, 7, 8}.
Now,
Number of elements in A and B as,
⇒ n(A) = 4 and,
⇒ n(B) = 8
Thus,
the number of one to one function from A into \[\text{B}={{8}_{{{\text{P}}_{4}}}}\]
\[=\dfrac{8!}{4!}\]
\[=\dfrac{8\times 7\times 6\times 5\times 4!}{4!}\]
After dividing, we get
$\Rightarrow 8 \times 7 \times 6 \times 5 \\
\Rightarrow 1680$
Note:
This type of question is counting based. For counting in maths we use permutation and combinations and factorials. factorial notation on $n$ is $n!$ and it means multiplication of natural numbers upto $n$.
Find the number of elements in A and B i.e. n(A) and n(B). Thus, no. of one-one function from A into B can be find by using \[{}^{n(B)}{P_{n(A)}}\]
Complete step by step solution:
Given,
A = {1, 3, 5, 7}
and, B = {1, 2, 3, 4, 5, 6, 7, 8}.
Now,
Number of elements in A and B as,
⇒ n(A) = 4 and,
⇒ n(B) = 8
Thus,
the number of one to one function from A into \[\text{B}={{8}_{{{\text{P}}_{4}}}}\]
\[=\dfrac{8!}{4!}\]
\[=\dfrac{8\times 7\times 6\times 5\times 4!}{4!}\]
After dividing, we get
$\Rightarrow 8 \times 7 \times 6 \times 5 \\
\Rightarrow 1680$
Note:
This type of question is counting based. For counting in maths we use permutation and combinations and factorials. factorial notation on $n$ is $n!$ and it means multiplication of natural numbers upto $n$.
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