How many functions are members of $F$?
I was asked a question like this. I've given the answer
$$\dfrac{m!}{(m - n)!}$$
But they said i was wrong and the answer is $mn$
Where am i wrong?
Please help. Thanks !
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$\begingroup$ Your answer is indeed correct. "They" probably missed the fact that F contains only one-one functions. If any functions were allowed, then "they" are right. $\endgroup$– Gautam ShenoyJan 14, 2014 at 4:58
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2 Answers
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We have $m$ choices for what $1$ is sent to. For each of these, we have $m-1$ choices for what $2$ is sent to. For each way of doing these two things, there are $m-2$ choices for what $3$ is sent to, and so on for a total of $$m(m-1)(m-2)\cdots(m-n+1).$$ This can also be written as $\frac{m!}{(m-n)!}$.
Your calculation is correct. The answer $mn$ is not, unless the problem was incorrectly described.
answered Jan 14, 2014 at 4:11
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Who is this "they"? You first pick the range of your function (can be done in $\binom{m}{n}$ ways, then any permutation thereof gives you a function (and a different one), giving your formula, and not "theirs".
answered Jan 14, 2014 at 4:10
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$\begingroup$ It's always "they" who create problems by dictating whether we are right or wrong regardless of our method. $\endgroup$ Jan 14, 2014 at 4:59
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