functions-combinations

Question

Assume an almost injective function is a function in which exactly two element from domain maps to a single element in co-domain, otherwise function is injective. $S$ and $R$ are sets with cardinality $m$ and $n$ respectively. $(m<n)$ and $m\geq 2$. Number of almost injective functions from $S$ to $R$ is _____________.
options

• A. $\frac{n!m(m-1)! } { [2(n-m+1)}$
• B. $\frac{n!m(m-1) } {2(n-m+1)}$
• C. $\frac{m!n(n-1)! }{ [2(n-m+1)!}$
• D. $\frac{( n-m)!}{ 2}$

put m=2 and n=5. we get almost injective functions=5
shudnt it be C?

Comments

options very less visible :(

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edited now

Answer 1

Exactly 2 elements in S will point to 1 element in R. Therefore, total combinations=> ^mC₂ * ⁿC₁ 

Rest (m-2) elements in S will point to exactly 1 of (n-1) elements in R. So, no of 1:1 functions => ^(n-1)P_(m-2)

Multiplying both results, we get =>

n! m(m-1) / 2* (n-m+1)!

So the answer must be B. There is a typo in that option.

Comments

But it says exactly $2$ from domain point to $1$ in other set. How can remaining element point to 1?

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Are you saying '1' as in the number 1 ?

Because the question is saying 'single element' and not 1 specifically.

 

Or if you are saying that exactly two elements will point to single elements, rest will point to multiple, then it is not even a function, isn't it?

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Ohh!! I understand my mistake, it says exactly 2. I considered it to be every pair of $2$

Answer 2

According to me, the answer should be n! m(m-1) / 2* (n-m+1)! (m-2)!

as selecting any 2 elements from “m”= mC2  and selecting any element from “n” for its mapping= nC1

so we get mC2 *nC1  ----(1)

now for the remaining “m-2” elements in “m” we can select them from “n-1” for mapping as n-1Cm-2 -----(2)

combining both (1) and (2)

(mC2) (nC1) (n-1Cm-2)  which should be the above one as mentioned.