Permutations of the set $\{1, 2, 3, 4, 5\}$ fulfilling certain conditions - Combinations Theory

Problem:

Among all the possible permutations of the set $\{1, 2, 3, 4, 5\}$, in how many fulfills that:

1. the element $1$ is in the first position?
2. the element $2$ is in the second position?
3. the first three elements occupy the first three positions?
4. Any of first three elements is not in their correct position?

What have I tried?

#1:

If the element $1$ remains in first position I see it as a permutation of the other elements which is:

$$4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$$

#2:

I see this problem as the same as the first question but just for another element so I believe it to be the same answer which is:

$$4! = 24$$

#3:

If the first three elements remained in their original positions, I see that as a permutations of the remaining 2 elements which appears to be:

$$2! = 2 \cdot 1 = 2$$

#4:

$\color{red}{\text{I am not sure how to correctly proceed with this question. }}$

But from my understanding, it would be a situation where:

• The first element remains in it's original position with permutations of the other element then the same for the second then the third element.

• Also, when the first two elements remain, then the second two and finally the first and third.

And would it be correct that it is the same as:

$$3(4!) + 3(3!) = 3\cdot24 + 3\cdot6 + 2 = 72 + 18 = 80$$

Would these be correct and if not where did I go wrong and how do I correct it?

• Concerning #3, it doesn't say that the first three must occupy their original positions, it says that the first three must be in the first three positions. $3!$ ways for three elements to be in three positions. Then $2!$ for the other two. Thus the answer is $3! \cdot 2!$. – Joel Dec 30 '17 at 22:03
• For #4, just observe that the opposite of "any of the first three is not in its position" is "all of the first three are in their position", – celtschk Dec 30 '17 at 22:07
• @celtschk Would that be the same as $5! - 2!$? – Omari Celestine Dec 30 '17 at 22:32
• @OmariCelestine: Yes. – celtschk Dec 31 '17 at 7:01

No. 3

It is said that the first 3 occupy first three places so 1, 2 and 3 can be arranged among themselves in 3! ways. So total number of ways these can be arranged = 2! x 3! = 12

No. 4 • I've read number 4 as "at least one of the first three is not in the correct position", while you apparently read it as "none of the first three is in the correct position". – celtschk Jan 16 '18 at 10:35

For #4, you can use Inclusion-Exclusion principle -

$|U| = 5!$ (The general case, "universe")

$|\bigcup \limits_{i=1}^n A_i|$ = $3\cdot4! - 3\cdot3! + 2!$ (we use it to avoid double counting)

By combining them we get,

$|U| - |\bigcup \limits_{i=1}^n A_i| = 5! - 3\cdot4! + 3\cdot3! - 2! = 64$

• Why subtract the $2!$? – Omari Celestine Dec 30 '17 at 23:11
• That's the Inclusion-Exclusion principle, we define $|\bigcup \limits_{i=1}^n A_i|$ as - $|M \cup P \cup C|=|M|+|P|+|C| -|M\cap P|-|M \cap C|-|P \cap C|+ |M \cap P \cap C|$ (M,C,P would be ball 1,2,3 not in their correct position) Thus, subtracting From the general case, $|U|$ we get - $|U| -(|M|+|P|+|C|) +|M\cap P|+|M \cap C|+|P \cap C|- |M \cap P \cap C|$ (the last one is 2!, hence we subtract it) I guess it would be better if you read the usage in Wikipedia, it's really simple. – Yariv Levy Dec 30 '17 at 23:14
• Given that there are already $4\cdot 4!=96>64$ permutations where the first element is not $1$, and those are a subset of those where any of the first three elements is not in its place, it is obvious that your calculation cannot be correct. – celtschk Dec 31 '17 at 7:10