number of function

Question

How many functions are there from the set {1, 2, . . . , n}, where n is a positive integer, to the set {0, 1}
a)  that assign 1 to exactly one of the positive integers less than n?

Comments

A small doubt understanding the question... Why 1 can be assigned to n ?

Isn't the question says that 1 can only be assigned to exactly one number between 1 to n-1 ?

Plz help

Answer 1

The domain set has $n$ elements and co-domain set has 2 elements. So, each of the $n$ elements from domain has 2 choices in the function and thus we get $2^n$ total functions.

Now, we are given a condition that exactly 1 positive integer less than $n$ maps to 1. So, all others less than $n$ must map to 0. We can find this number in $^{n-1}C_1$ ways (all the mappings for these $n-1$ elements are fixed)  and $n$ can be mapped to either 0 or 1, thus we get $2. (n-1)$ possible functions.

Comments

sir,suppose set A={1,2,3} and given set {0,1}!!

acc to condition 1,2 from set A can be mapped to 1 because they said +ve no less n  but only one at a time from SET A... means if 2 from set A mapped to 1 in set {0,1} than reamaining elements from SET A ie 1,3 will be mapped to zero... so total we should get only 2 functions from this example..

how 2(n-1).??

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In your example, you mapped 2 ->1, now 1->1 can also be made and we get 2 functions. Here, $n$ is 3, and for the $n^{th}$ element mapping can be to either 0 or 1. So, 3 can be mapped to either 0 or 1 and thus we get 2 + 2 = 4 possible functions.

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thanks i got now:)

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arjun sir why r u multiplying  plz explain..

it should be (n-1) + 2.. plz correct me if iam wrong

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but in this way overall 1 is getting assigned two times .................

Answer 2

according to me answer is as out of n-1 elwments only is mapped to 1 sp remaining n-2 elements + nth element make it n-1 element which will be mapped to 0 so n-1C1 +n-1C1 =2(n-1)C1 so 2(n-1)

Comments

No, you are doing wrong. Because the question allows $n$ to be mapped to either 0 or 1. Also, you select one out of $n-1$ elements first. Then how you get ${}^{n-1}C_1$ again?