I am super confused with permutations and combinations. I just don't see when you are supposed to use each one, and when you aren't supposed to use either.
Our homework was to correct our tests, and I did quite poorly. I tried to fix some of these, so could someone check my answers?
1. In how many ways can 10 people be seated around a circular table?
I really don't get this. On one of homework problems, there was a question like this and the answer was 1 way. I put that on my test, and it was wrong. The only other way I could think to do this was 10! (10 factorial) or 3628800
When would there be only 1 way?
2. There are 6 red bean bags, 5 blue bean bags, and 3 green bean bags in a container. How many ways can 4 bean bags be selected so that:
A) all 4 are red?
B) 2 are blue and 2 are green?
C)exactly 3 are red?
I don't even know how this one would be solved..
3. A license plate for a certain state has 4 digits followed by 2 letters. How many different plates are there if the letters may be repeated but the digits may not?
The only way I could think of to do this would be:
P(6,4)C(6,2)=360 x 15=5400
And that was the wrong answer..
4. Using only the letters {R, S, T, U, A, B, C, D, E}, how many 6-letter codes are possible in which the following is true?
A) Letters can't be repeated.
B) Letters can be repeated.
C) The first letter is a vowel and Letters can't be repeated.
D) The first letter is a vowel and Letters can be repeated.
I thought the answers would be C(9,6), P(9,6), C(8,6), P(8,6) but I got them all wrong. How do you do this?
It would really help me if someone would check these, or show me how to do some of them.
It would really help me if someone could check these. Please?
In a very general explanation,
in a permutation the order or the position of your arrangement matters,
while in a combination the order does not matter.
e.g. Suppose from a class of 25 students, 6 are to be chosen to form a mathteam
The order in which they are chosen does not matter, so that would be a combination.
On the other hand if those 6 students are to become experts in different areas of math on the team, that would be a permutation.
1. This is one of those special cases.
I will asssume that there is no specific order or special chairs to be filled.
So let's fill the first chair, that would be 10 ways, the second in 9 ways, etc.
so that would be 10! ways, right?
Now suppose everybody got up and moved to the chair on their left, and then sat down. Would the arrangement of people sitting around the table have changed? NO!
Now think of how many ways they could do this? Could they not do this 10 times, each time resulting in the original arrangement?
So the 10! ways has to be divided by 10, giving you 9! ways.
3. for the license plate digits may not be repeated, but letters may.
Can the plate start with a 0? I will assume it can.
so the first 4 numbers are 10*9*8*7
followed by 2 letters which would be 26*26
so the number of plates is 10*9*8*7*26*26
4. you have 9 different letters, of those arrange 6, (the order matters)
a) the number of ways would be 9*8*7*6*5*4
b) they can be repeated, so 9*9*9*9*9*9
or 9^6
c) 3 ways to fill the first spot with a vowel, after that anything goes without repeating the letters
so no of ways = 3*8*7*6*5*4
d) only change is 3*9*9*9*9*9
Let me think about #2, it seems like it should be a probability question
which would be P(9,6)
the very last line of "which would be P(9,6)"
should have come after 4.a) to read
" a) the number of ways would be 9*8*7*6*5*4
which would be P(9,6)"
2. I am a bit mystified by the wording.
They are using the word "selected" which suggests a combination, in other words, the order does not matter
in a) there is only one way for 4 bags to be red, namely RRRR
b) by the same thinking there is only one way for 4 bags to be 2 blue and 2 red,
BBRR (BRRB would be the same since order does not matter)
c) exactly 3 of the 4 bags must be red, meaning that the 4th could be either blue or green
which would be 2 ways: RRRB or RRRG
Oh ok, I get it. But i have one more question.
Why are there always more ways when you use a permutation? Wouldn't you think that you would have more ways when you used a combination since the order doesn't actually matter? I don't know if that makes sense to you. It's kind of bugging me
suppose from you 5 friends you "choose" 3 to ride in your car.
That would be C(5,3) = 10
now suppose you pick 3 people to sit in the passenger seat, the rightrear seat and the leftrearseat.
that would be 5*4*3 or P(5,3) = 60 ways
notice the permutation is more.
Another way to look at it:
suppose you pick one of the 10 combinations from my example.
once you have selected the 3 people you want you could now fill the
frontseat, rearleft and rearright seats in 3*2*1 or 6 ways
And of course we could do these 6 arrangements for each of our 10 choices, giving us a total of 60 arrangements.
(I am still puzzled by #2. Are you studying probability questions yet?)