Online Course Discussion Forum

Self Paced 1A Week 5 Advanced Counting

 
 
Picture of Kathy Liu
Self Paced 1A Week 5 Advanced Counting
by Kathy Liu - Thursday, 11 January 2018, 7:12 PM
 

Hi Im a bit stuck on the reasoning behind some of the ways a few of the problems are being solved. 

For F6 the question is : Supposed Alice, Bob, Charles, and Desiree are 4 students comparing the days of the week on which they were born. How many possible outcomes are there?

A: 4^7

My question is why cant this be 7^4 because each person can be assigned any one of 7 days so why wouldnt it be 7*7*7*7?


For F8 and F9

8: Recall F6. How many possible outcomes are there if they were all born on different days? Assume we dont care about who was born on a given day.

A: 7 choose 4

9: Recall F6. How many possible outcomes are there if they were all born on different days? Assume we now care about who was born on a given day.

A: 7!/(7-4)! 

My question is why cant these two be swapped, the solution on the answers isn't clear enough. I understand that one is without repetition and without regard to order while the other is without rep but respecting order but  how is it so and what is the reasoning behind why it was solved the way it is? Can they be swapped?

 
Picture of David Reynoso
Re: Self Paced 1A Week 5 Advanced Counting
by David Reynoso - Tuesday, 16 January 2018, 1:56 PM
 

You are correct. The answer for F6 should be $7^4$, not  $4^7$.

On F8, we do only care which days someone was born in, but it makes no difference which of them was born on each day. Thus, we only need to select $4$ different days out of the $7$ days of the week., that is, $\binom{7}{4}$.

On F9 we do care about who was born on each day, and since each person was born on a different day, we can use permutations to figure that out. Say you first choose which of the $7$ days Alice was born, then which of the $6$ remaining days Bob was born, and so on.  That is, $7\times 6\times 5\times 4$, or $7P4 = \frac{7!}{3!}$.