Craig is traning for a race. He bikes every 2 days and swims every 3 days. If he biked and swam today, how many days will pass before he bikes and swims on the same day again, the least common multiple of the numbers of days?
Multiples of 2: 2,4,6,8,10,12, ...
Multiples of 3: 3,6,9,12,15,18, ...
Answer
Verified
Hint: In this problem, we have to find how many permutations of the letters “a b c d e f g h” contain. Permutation determines the number of techniques that determines the number of possible arrangements in a set when the order of arrangement matters. We can also use formulas to find the answer. In this problem, we have a group of letters “a b c d e f g h”, which has 8 letters in it, we can either directly find the factorial for the given number of letters or we can use the permutation formula to find the answer.
Complete step by step solution:
We know that the given letters are “a b c d e f g h”.
Where there are 8 letters, n = 8, r = 8.
We know that the permutation formula is
Permutation, \[^{n}{{P}_{r}}=\dfrac{n!}{\left[ n-r \right]!}\]
Where, n = 8, r = 8.
\[{{\Rightarrow }^{8}}{{P}_{8}}=\dfrac{8!}{\left[ 8-8 \right]!}=\dfrac{8!}{1}=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1=40320\]
Therefore, there are 40320 permutation in the letters “a b c d e f g h”.
Note: We can also find the given permutation, as it has 8 letters in it, so we can just find the factorial for the given number of terms to find the factor.
We know that the given letters are “a b c d e f g h”.
We can see that there are 8 letters in it, so we can find the factorial of 8
\[\Rightarrow 8!=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1=40320\]
Therefore, there are 40320 permutation in the letters
“a b c d e f g h”
No, EBCD is not a permutation of ALL of the letters; it is a permutation of only 4 of the 7 letters. The question asks for permutations of the entire string.
So, out of the 7! = 5040 ways to arrange the 7 letters, 5! = 120 contain BCD together, in that order.
This is a simplistic way to think of it. If you try to mechanically equate each item in the string to a number, you will make mistakes in any problem harder than the very simplest.
We are not simply assigning a number to each item; we are counting ways to choose. There are 5 ways to choose the first, 4 ways to choose the second, and so on. So, if anything, I would "assign" the numbers in the opposite order. But they are assigned to choices, not to items.
Watch out for words here! These are not "combinations", which is a technical term meaning subsets [without ordering]. They are permutations, or orderings.
And the idea is that there are 2 choices for the first, and 1 for the second [because we don't allow repetitions].
There is no difference at all between arranging BCD and A, and arranging BCD and E.
It is becoming even more clear that your faulty model is leading to your confusion.
You have so far counted the permutations of 3 items "taken 2 at a time", which is written as 3P2. These are not part of the permutations of all 3 [since these only include 2], but you could continue from here if you were careful.
At this point, your main error appears to be that you think the permutations you were asked to count don't have to include all items.
Are you taking a course, or reading a book? If so, tell us what it is, so we can have a better idea where you are coming from. If not, you really need to find one, because starting from the beginning, learning what each term means and how each concept is to be thought about, would help a lot.
I'm studying and reading through my discrete math book.. I seemed to be grasping the idea of permutations, but I don't understand how the solution for this particular problem came to be.
Question:
How many permutations of the letters ABCDEFGH contain the string ABC ?
Solution: Because the letters ABC must occur as a block, we can find the answer by finding the number of permutations of six objects, namely, the block ABC and the individual letters D, E, F , G, and H . Because these six objects can occur in any order, there are 6! = 720 permutations of the letters ABCDEFGH in which ABC occurs as a block.
My reasoning?
I know the permutation formula is: $\frac{n!}{[n-r]!}$ and there are 8 letters. We want ABC, so r=3? Using this formula.. wouldn't you get: $\frac{8!}{[8-3]!}$ = $\frac{8!}{5!}$ = 336?
I guess since ABC is a "block" then ABC is kind of like a single letter? so 8-3=5.. then +1 for the [abc] combo? Like I think I get that after looking at the solution, but If the solution wasn't provided I don't understand how you would know to approach it that way.
I'm probably just lacking common sense and this is a dumb question, but is this just something that should be common sense?
Thanks