Statistics and probability
How many ways can we pick officers for our organization? Created by Sal Khan and Monterey Institute for Technology and Education.
Want to join the conversation?
- What's the intuition for something like this:
How many unique ways can you arrange the letters in the word HAPPY?
Well, if we had left out the word unique in the problem above, the answer would be 5! or 120 (at least if I know what I'm talking about).
But in our potential permutations, we would have duplicates---the two P's are used interchangeably.
So you divide by 2! based on the information I gleaned from the hints on the exercise "Permutations and Combinations".
Why is that though, when ignoring the formulas and thinking purely about the theory of it?
If you could relate the above problem to another I've already learned about in one of these last few videos, that would be great. If not, it doesn't really matter.
Please use the problem I came up with as an example.(8 votes)
- There is no need to relate to previous videos, you state all the information needed to answer your question. Using the word "HAPPY" (let's distinguish the P's as P1 and P2). One arrangement of the letters is, of course: H A P1 P2 Y
Another is: H A P2 P1 Y
But in practice, since P1 and P2 cannot be distinguished, these two are equivalent arrangements of the letters. When we start with 5!, we are over-counting: the factorial doesn't realize that the two P's are identical, it treats all 5 letters as distinct. Therefore, for every single arrangement of the 5 letters, we can make another identical one by switching the P's. So when we compute 5!, so know that we need to divide that number by 2 to account for switching the P's about. Hence, 5!/2 is the number of unique ways to arrange the letters of HAPPY.
The general idea is that once we count the number of ways to arrange all letters (treated as being distinct), we need to divide by the number of ways to arrange the repeated letters. In this case, it's 2 because 2!=2. If we had three repeated letters (e.g. "TITTER"), then we'd divide by 3!, because there are 3! ways to arrange T1, T2, and T3.(21 votes)
- This problem would be a permutation, right?(6 votes)
- That is exactly it! This is a good example of a permutation. He doesn't show the permutation equation, but yes, that's exactly what's going on.(10 votes)
- Would this be a permutation or combination problem, if you were sticking strictly to the formulas?(5 votes)
- In this case order of choosing the people matters so it will be a permutation.
The first person is the president , the 2nd is the Vice President and the 3rd is the Secretary so it matters if the person is 1st,2nd or 3rd. If they were being selected for the same positions then order would not matter and it should be a combination. Always, if the order of selecting matters it is a permutation and if it does not it is a combination.(5 votes)
- What do you mean by right numbers? The numbers that are in what ever problem you do? Or I am wondering something right now. Why is it that addition and multiplication, you do not care about the order, and that division and subtraction, you do care about the order? Is it because adding is like positive, and subtracting is like negative? Oh and do two negatives cancel out to be a positive?(4 votes)
- 1. And 2.) The right numbers are the numbers that are used in the problem. In this case, they are 9, 8, and 7.
3.) Great question. Look below.
As you can see, you CAN switch the numbers in subtraction; you just have to make sure you carry the minus sign with the number right of it when you switch the numbers.
Now look below again.
As you can see, you also CAN switch the numbers in division; you just have to remember to carry the division sign with the number right of it when you switch the numbers.
If you forget to carry the signs...
Above, we took the negative, or additive inverse, of the first problem when trying to switch the numbers without the signs.
Above, we took the reciprocal, or multiplicative inverse, of the first problem when trying to switch the numbers without switching the signs.
I hope this successfully answers your question!(3 votes)
- How many ways are there to choose a five letter word from the letters in the word "INDEPENDENT". Please give me an answer to this. It is an IITJEE question.(4 votes)
- There are 11 letters, so the number of permutations is 11!
Since the order of equal letters don't matter, divide by 2! for the two Ds. Divide by 3! for the three Es. And so on.(2 votes)
- Does someone could explain this question
You need to put your reindeer, Jebediah, Lancer, Ezekiel, Gloopin, and Bloopin, in a single-file line to pull your sleigh. However, Lancer and Ezekiel are best friends, so you have to put them next to each other, or they won't fly.How the result is 48?(2 votes)
- Lancer and Ezekiel must be next to each other, so we can consider Lancer-Ezekiel to be one reindeer for a moment.
Then there are four reindeer to place into four slots. We have 4 options for the first slot. For each of those four options, we can put one of three reindeer in the second slot, for a total of 12. Then, for each of those 12, we can put one of two reindeer in the third slot, for a total of 24. Then there's one reindeer for one slot, so this doesn't give us new choices. We can order them 24 ways.
However, remember that we lumped Lancer and Ezekiel into one reindeer. In each of these 24 configurations, we could put Lancer first or Ezekiel first. So this doubles the number of configurations again, for a grand total of 48.(4 votes)
- Q-Four different items have to be placed in three different boxes. In how many ways can it be done such that any box can have any number of items?(2 votes)
- This is an example of a ball and urn problem. The answer is 6 C 2, and ill explain why: each ball is an X and each box is a "space". there are __|____|__ 3 boxes. in each box can be placed 4 balls, so like the comment above, this can be any number of possibilities. for example, XX|__X_|X_. There are 2 lines and 4 balls, and youre choosing 4 balls to go between the 2 lines, so the answer is 6 C 2, or 15(3 votes)
- If I have Five People, and Five Officer Positions, then the number of permutations is 5!, which equals 120 possible permutations. However, if I have Five People and only Four officer positions, then the number is 5! / 1! the answer is also 120 possible permutations. How can rectify this logically?(1 vote)
- When picking the first four (assuming order), there are 120 possibilities. The last person is stuck into the last position no matter what, there's no real choice for him. This is true whether it's an officer position, or whether it's "no position."(6 votes)
- This is bad example. The order doesn't matter given that the questions is, how many ways can you choose THE BOARD. If I pick Sam, Jane, and Tom for Pres, VP and sec, that is the same "board" as one in which Tom, Sam, and Jane are Pres, VP and secretary. i.e., the members of the board are the same in either case regardless of the positions they hold. Isn't the problem you are solving the one in which you ask, "out of nine board members, how many ways can you assign 3 of the members to the positions of Pres, VP, and secretary?"(3 votes)
- I definitely agree that it is worded weirdly. However, seeing that it's in the permutation category, it's assumed that it is solved this way.(2 votes)
A club of nine people wants to choose a board of three officers: a President, a Vice President, and a Secretary. How many ways are there to choose the board from the nine people? Now, we're going to assume that one person can't hold more than one office. That if I'm picked for President, then I'm no longer a valid person for Vice President or Secretary. So let's just think about the three different positions. So you have the President, you have the Vice President, VP, and then you have the Secretary. Now, let's say that we go for the President first. It actually doesn't matter. Let's say we were picking the President slot first and we haven't appointed any other slots yet. How many possibilities are there for President? Well, the club has nine people, so there's nine possibilities for President. There's going to be nine possibilities for President. Now, we're going to pick one of those nine. We're going to kind of take them out of the running for the other two offices. Right? Because someone's going to be President, so one of the nine is going to be President. There's nine possibilities, but one of the nine is going to be President. So you take that person aside. He or she is now the President. How many people are left to be Vice President? Well, now there's only eight possible candidates for Vice President? Eight possibilities. Now he or she also goes aside. Now how many people are left for Secretary? Well, now there's only seven possibilities for Secretary. So if you want to think about all of the different ways there are to choose a board from the nine people, there's the 9 for President times the 8 for Vice President times the 7 for Secretary. You didn't have to do it this way. It could've been-- you could've picked Secretary first. Then there would've been nine choices. And then you could've picked Vice President, and there would've still been eight choices. And then you could've picked President last, and there would've only been seven choices. But either way you would've got 9 times 8 times 7. And that is, let's see, 9 times 8 is 72 times 7 is-- 2 times 7 is 14, 7 times 7 is 49 plus 1 is 50. So 504 possible ways to pick your board out of a club of only nine people.