How many functions f from {−1005, . . . , 1005} to {−2010, . . . , 2010} are there such that the following
two conditions are satisfied?
• If a < b then f(a) < f(b).
• There is no n in {−1005, . . . , 1005} such that |f(n)| = |n|.
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1 Answers
1i m posting the solution,i have not understood it,.
so,can anyone explain me now!!
The
1st condition tells us that f is strictly increasing, and the second condition
tells us that f(n) is not equal to '' n" or "-n"for all n in the domain. We defi
ne another function g with g(n) = f(n) + 1 if
f(n) < n, g(n) = f(n)-1 if f(n) > n, and g(n) = f(n) otherwise. So g(n) is any increasing function
with domain {1005;1004; : : : ; 1005} and range {2009;2008; : : : ; 2009}, but not necessarily with
g(n) equal n or -n. Since the mapping between g and f is unique, it is a bijection, so the number of such
g is equal to the number of such f. The number of such g is just the number of strictly increasing
sequences with 2011 terms with 4019 possibilities for each term,
so (4019 C 2011) is the answer