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Instead of counting directly, which would involve messy casework, we will count the complement of what we want (how many functions $f$ are one-to-one) and then subtract the result from the total number of functions.

If $f$ is one-to-one, then there are $5$ choices for the value of $f(a)$ (any of the elements of $Y$ ), there are $4$ choices for the value of $f(b)$ (it can't equal $f(a)$ ), there are $3$ choices for the value of $f(c)$ (it can't equal $f(a)$ or $f(b)$ ), and there are $2$ choices for the value of $f(d)$ (it can't equal $f(a),$ $f(b),$ or $f(c)$ ). Thus, there are a total of $5\cdot4\cdot3\cdot2 = 120$ one-to-one functions $f.$

If we ignore all restrictions, then for each of the $4$ values in the domain of $f,$ there are $5$ choices for the output. So, there are $5^4 = 625$ functions, counting without restrictions.

Since every function is either one-to-one or not one-to-one, there are exactly $625 - 120 = \boxed{505}$ functions that are not one-to-one.

Each of the element of $X$ has 5 choices,to make up all functions. Now 1-1 functions will be those in which 4 out 5 elements are chosen from $Y$ & connected distinctly by elements of $X$ , which gives $^5P_4$ ways. So number of required non 1-1 functions are $5^4 - ^5P_4 =505$ in number.

Instead of directly trying to count what is asked, it is easier to count the number of injective maps and the total number of maps. One can then subtract the former from the latter. To get the total number of maps, we notice that each of the elements of $X$ has five possible images, giving $5^4=625$ total maps. The number of injective ones is $5\cdot4\cdot3\cdot2=120$ : $a$ has $5$ possible images, after which $b$ has only $4$ options left, etc. The answer thus is $625-120=505$ .

For a function to be one-to-one, each element in Y that is in the range, corresponds to exactly only one element in the domain X .

First we find total mapping possibilities for the function.

$5*5*5*5 = 625$

In this case there are no restriction, so each element in X has 5 choices to map to in Y .

Next we find number of possibilities were the function is one-to-one.

$5*4*3*2 = 120$

In this case, the first element in X has 5 choices to map to, then each succeeding element has 1 less choice since only 1 element in X can map to an element in Y .

Once we know the total number of possibilities and the number that are one-to-one. We simple subtract the latter to find the number that are not one-to-one.

$625-120 = \fbox{505}$

If the function is not one-to-one, then two values from $X$ must go to the same value in $Y$ .

We count these using complementary counting. The total number of functions is $5^4=625$ , since each value in $X$ can go to any value in $Y$ . If the function is one-to-one, then the values must be distinct. There are $\dbinom{5}{4}=5$ ways to choose 4 distinct values in $Y$ , and $4!=24$ ways to map $X$ to this set of 4 values.

The answer is $5^4-\dbinom{5}{4}4!=625-120=\boxed{505}$ .

Total number of functions possible is $5^4$ because for each element in $X$ , we can choose from 5 elements in $Y$ . The total number of one-to-one functions is: $5*4*3*2$ since for the first element in $X$ , we can choose an element in $Y$ in 5 ways but for the next element in $X$ , we are allowed to choose from only 4 elements in $Y$ and so on. So the total number of functions that are not one-to-one are $5^4-5*4*3*2=505$ .

we've not one-to-one function, is the function that break injectivity, because it's a function than the order of elements matter, than we've $5^4$ functions than can be. (the number of combinations of the 4 elements from 5)

but we need all functions that break injectivity so: the number of injective functions is $\binom{5}{4}\times4!$ .

why $4!$ ? is the number of permutation of each set of $4$ elements because the order matter, than:

$5^4 - \binom{5}{4}\times4! = 505$

Because X have 4 members and Y have five members, the number of whole functions is 5^4. But, the one to one function is 5!. So, the 'not one to one' function can be found from the rest: 5^4-5!= 625-120= 505

5^4-5!=505 gg

= Total possible functions - One-to-one Functions

= $5^{4} - 5P4$ = 505

First, we count the total number of functions from set $X$ to set $Y$ . To do this, note that each element of $Y$ gets mapped to exactly one of the four elements of $X$ . Thus each of the $5$ elements belonging to $Y$ is faced with $4$ choices. By the multiplication principle, there are $5^4= 625$ ways to do this.

Now we count the total number of one-to-one functions from $X$ to $Y$ . First note that there are $5$ elements in $Y$ and $4$ elements in $X$ . So one of the elements in $Y$ is not mapped with any of the elements in $X$ . We take any $4$ elements from $Y$ , permute them, and assign the first element of the permutation to $a$ (i.e $f(a)= 1^{st}$ element of the permutation), the second element of the permutation to $b$ , the third element of the permutation to $c$ , and the fourth element of the permutation to $d$ . Note that we have a bijection here: each such permutation corresponds to a unique function, and each function corresponds to a unique permutation. There are $^5P_4$ ways to do this.

To find the number of functions which are not one-one, we have to subtract this number from the total number of functions. Hence our answer will be $5^4 - ^5P_4= \boxed{505}$ .