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Math Challenge IC Handout 8

 
 
Picture of Christina Peng
Math Challenge IC Handout 8
by Christina Peng - Friday, 27 April 2018, 12:27 AM
 

What does bijection really mean? I feel like from the lesson that it means there are the same amount results/ ways from both sets. Is all I need to do for the bijection questions making sure they have the same amount of results?

8.17 to 8.20: I don't know what to do except for 18 where I used stars and bars

8.21 : Your goal is to pass THROUGH the letter C right?

8.22: How does the answers connect? The problem doesn't seem to and my answer for 8.22 is way bigger than the answer for the earlier problem. Did I do it wrong?

8.23 How would a function or sequence look when you try to wrrite it out like that? Is changing f(x) to t(n) all you do?

8.27 I have no idea how to start.

8.30: what does it mean to not stand at the ends? Is it that none of them can stand at the end or one of them could while the other doesn't?


Sorry about so many questions, I'm just really really really confused with this unit. Thanks.

 
Picture of David Reynoso
Re: Math Challenge IC Handout 8
by David Reynoso - Monday, 30 April 2018, 11:40 AM
 

For 17, 18, 19 and 20 try to identify first if you care about order and if repetitions are allowed. On 18 it is OK to use stars and bars since repetitions are allowed and we do not care about the order.

On 21, yes, you want the path to go through the letter C.

On 22 there should be $5$ possible ways of filling in the table as well. For each of the tables there is one path that can be associated with the table. Try comparing the $5$ paths with the $5$ tables.

For 23, remember that a bijection is a rule that matches the elements of two sets (in this case one is the set of all functions from $A$ to $B$ and the other is the set of sequences of length $4$ made up of $0$'s and $1$'s. So, we want to find a rule that tells us what sequence to write if we have a function from $A$ to $B$.

On 27 it might be good to start by looking at $6 = 1+1+1+1+1+1$. Any other possible sum could be obtained by removing some of the $+$ signs. For example you can think of $6 = 1+2+1+1+1$ as $6 = 1 + 1 \not + 1 + 1 + 1 + 1$.

On 30, none of them could stand at the ends (so the first and last persons in line are some of their friends).

Picture of Patrick Zhang
Re: Math Challenge IC Handout 8
by Patrick Zhang - Tuesday, 1 May 2018, 4:54 PM
 

Hey, I have a question:

For 8.15, is it considering repeats but in a different order as a completely different sum? For example: 4+1+1 is different from 1+4+1?

Picture of Christina Peng
Re: Math Challenge IC Handout 8
by Christina Peng - Tuesday, 1 May 2018, 9:22 PM
 

I got 8!/(6!*2!) so I think you count them differently?

Picture of Christina Peng
Re: Math Challenge IC Handout 8
by Christina Peng - Tuesday, 1 May 2018, 9:31 PM
 

Thank you so much!