Not bijections

Loki123 said:
I have no idea how to solve this. Any advice? View attachment 32528
Click to expand...
1. is right.

2. You need to rethink this. For just two elements x, y:
{(a, x), (b, x), (c, x), (d, y)}
{(a, x), (b, x), (c, y), (d, x)}
{(a, x), (b, y), (c, x), (d, x)}
{(a, y), (b, x), (c, x), (d, x)}

{(a, x), (b, x), (c, y), (d, y)}
{(a, x), (b, y), (c, x), (d, y)}
{(a, y), (b, x), (c, x), (d, y)}
{(a, x), (b, y), (c, y), (d, x)}
{(a, y), (b, x), (c, y), (d, x)}
{(a, y), (b, y), (c, x), (d, x)}

{(a, x), (b, y), (c, y), (d, y)}
{(a, y), (b, x), (c, y), (d, y)}
{(a, y), (b, y), (c, x), (d, y)}
{(a, y), (b, y), (c, y), (d, x)}

And that's just using x, y.

3. See 2.

4. Can you find any non-bijections here? Write them out if you aren't sure.

See if you can find another way to count these.

-Dan
 
pka said:
@Loki123 Can you tell us how many functions [imath]f:\{a,b,c,d\}\to\{a,b,c,d\}[/imath] are there?
How many of those are injections? Is every one of those injections a bijection?
Please answer.
Click to expand...
You can also figure out the total number of constant functions.
 
pka said:
@Loki123 Can you tell us how many functions [imath]f:\{a,b,c,d\}\to\{a,b,c,d\}[/imath] are there?
How many of those are injections? Is every one of those injections a bijection?
Please answer.
Click to expand...
i am having difficulty understanding how to show functions using permutations, combinations or variations
 
Loki123 said:
i am having difficulty understanding how to show functions using permutations, combinations or variations
Click to expand...
The reason you are having trouble is simply because those are not the way to do this question.
If [imath]A[/imath] is a set then [imath]|A|[/imath] is the number members in the set.
The number of functions [imath]f:A\to B[/imath] is [imath]|B|^{|A|}[/imath] If [imath]A=\{a,b,c,d[/imath]
The number of functions [imath]f:A\to A[/imath] is [imath]4^{4}=256[/imath].
The number of injections [imath]f:A\to A[/imath] is [imath]4!=24[/imath]. In this case every injection is a surjection.
The number of constant functions [imath]f:A\to A[/imath] is [imath]4[/imath]. The image set is only one term.
Do you see there are no permutations, combinations or variations to it?
 
pka said:
The reason you are having trouble is simply because those are not the way to do this question.
If [imath]A[/imath] is a set then [imath]|A|[/imath] is the number members in the set.
The number of functions [imath]f:A\to B[/imath] is [imath]|B|^{|A|}[/imath] If [imath]A=\{a,b,c,d[/imath]
The number of functions [imath]f:A\to A[/imath] is [imath]4^{4}=256[/imath].
The number of injections [imath]f:A\to A[/imath] is [imath]4!=24[/imath]. In this case every injection is a surjection.
The number of constant functions [imath]f:A\to A[/imath] is [imath]4[/imath]. The image set is only one term.
Do you see there are no permutations, combinations or variations to it?
Click to expand...
YES! Thank you sooo much!
just one question, I don't understand what a constant function would represent here.
 
pka said:
Tectonically it is that the finial set is only one term. Every element of the set is mapped to the same term.
Click to expand...
So a, b, c, d would all be mapped into a, then b, then c, then d. Got it.
The question says which are not bijections or constant functions. So would the answer be 256-24-4=228?
 
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