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Let $A = \{2,4,6,8,10,12\}$ and $B = \{3,7,11\}$. Then no. of function such that element $3$

of $B$ has two , $7$ of $B$ has three and $11$ of $B$ has one pre - image in $B$ ,is

My Try:: for element $3$ of $B$.

we have to choose two element in $A$ out of total six elements.

This can be done by $\displaystyle \binom{6}{2} = 15 \;\; $ways

Similarly for element $7$ of $B$.

we have to choose three element in $A$ out of total four elements.

This can be done by $\displaystyle \binom{4}{3} = 4 \;\; $ways

Similarly for element $11$ of $B$.

we have to choose one element in $A$ out of total one elements.

This can be done by $\displaystyle \binom{1}{1} = 1 \;\; $ways

So Total ways $ = 15 \times 4 \times \times 1 = 60$

can anyone explain me i have done right or not if not then how can i proceed

Thanks

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  • $\begingroup$ The reasoning is right. You should say preimage in $A$, there is a typo, you wrote $B$. $\endgroup$ – André Nicolas Sep 17 '13 at 4:05
  • $\begingroup$ Thanks André Nicolas ......... $\endgroup$ – juantheron Sep 17 '13 at 4:13
  • $\begingroup$ You are welcome. It seems pointless to write an answer that says, yes, you are right. In writing up this sort of thing, it would be good if you said that for every choice of preimage for $3$, there are $\binom{4}{3} \dots$ to make he logic clear. And maybe at the end say that now the preimage of $11$ is determined, multiplying by $\binom{1}{1}$ seems mechanical. $\endgroup$ – André Nicolas Sep 17 '13 at 4:21
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Quoth André Nicolas:

Yes, you are right. In writing up this sort of thing, it would be good if you said that for every choice of preimage for $3$, there are $4\choose 3$$\ldots$ to make the logic clear. And maybe at the end say that now the preimage of $11$ is determined, multiplying by $1\choose 1$ seems mechanical.

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