1) If we have 5 letters: A, A, B, C, D, in how many ways (arrangements) can we form a 3-letter "word"?

How can I calculate this? The 2 A's seem to make things very complicated...and I have no clue how to do it...

Thanks for helping!:)

1) If we have 5 letters: A, A, B, C, D, in how many ways (arrangements) can we form a 3-letter "word"?

How can I calculate this? The 2 A's seem to make things very complicated...and I have no clue how to do it...

Thanks for helping!:)

Use the basic approach for solving for combinations:

Choices for the first 3-letter word = 5
for the second = 4
for the third = 3

then: 5 x 4 x 3 = 60. The total number of different combinations available.

But in this case, it's arrangements, so permutation instead of combination, right?

Also, there are two identical "A"s, so some arrangements will be double counted or so...

Kind of depends on exactly how the problem is worded. If they want to know how many distinct 3 letter words can be formed from AABCD then 4*3*2=60. If they want to know how many ways you can form those 60 words then you need to double the count on those words that use A. Which I think would simply be 60/4=15, so the total would be 75.

Denote 1st A=A1 and 2nd A=A2
A1BC and A2BC are counted as 1 "word" because actually A1=A2
Then how can I calculate the number of possible "words" now?


And also 4 x 3 x 2 = 24 isn't it?

Sorry, yes 4x3x2=24. It's a weekend. ;)

On further reflection though I think that Read-Only is right, or at least more right than I. The question is a bit vague to me. For example, is AAB a valid word?

I've managed to confuse myself now!

Sorry, yes 4x3x2=24. It's a weekend. ;)

On further reflection though I think that Read-Only is right, or at least more right than I. The question is a bit vague to me. For example, is AAB a valid word?

I've managed to confuse myself now!

Yep, the method I described is correct. So is the answer (60).;)

A1BC and A2BC are not the same word. You could compare it to using people and having two guys named John. Even though the names are the same they certainly aren't the same person. And that applies equally to the two "A"s.

Yeah, if you consider A1A2B and A2A1B to be distinct words, then clearly 60 is right. What leaves me uncertain is whether that is what the question as worded implies.

I tend to be overly critical of textbook questions. A lot of them seem to me to be poorly framed.

Yeah, if you consider A1A2B and A2A1B to be distinct words, then clearly 60 is right. What leaves me uncertain is whether that is what the question as worded implies.

I tend to be overly critical of textbook questions. A lot of them seem to me to be poorly framed.

Agreed. Far too many of them are more ambigious than the writer understands. He/she knows what THEY mean and ASSUME the reader will also. Poor writing skills, poor quality.

I think the answer is 33.

There are 9 cases which have 2 As.
AAB
ABA
BAA
AAC
ACA
CAA
AAD
ADA
DAA

There are 24 cases (4*3*2) for the rest of the possible words, some including only 1 A.
So, 24+9 = 33.

It's 60 words only if all of the letters are different.

A1BC and A2BC are not the same word. You could compare it to using people and having two guys named John. Even though the names are the same they certainly aren't the same person. And that applies equally to the two "A"s.
I think it's unlikely that this is what the textbook author meant.
If the question was supposed to simply be about finding permutations of three items from a set of five, then why would they list A twice?

I suspect they intend the student to count the number of permutations with both A's, then subtract half of that number from the 60 total permutations, leaving 51.

I think it's unlikely that this is what the textbook author meant.
If the question was supposed to simply be about finding permutations of three items from a set of five, then why would they list A twice?

I suspect they intend the student to count the number of permutations with both A's, then subtract half of that number from the 60 total permutations, leaving 51.

That might be, Pete - which takes us right back to what Kevin and I were saying. Also, the clincher may well be in something that none of here can see: an example or discussion in the textbook. We've all been having to work half-blind.

Well, it's kingwinner's text and question, so I think he gets the final say regarding question interpretation:
Denote 1st A=A1 and 2nd A=A2
A1BC and A2BC are counted as 1 "word" because actually A1=A2
Then how can I calculate the number of possible "words" now?

I suspect they intend the student to count the number of permutations with both A's, then subtract half of that number from the 60 total permutations, leaving 51.

You're still counting A1BC and A2BC as two different words. Since Zero already answered kingwinner's homework (STOP THAT!), here is how to rectify Zeno's and Pete's approaches.

Temporarily labeling the two As as A1 and A2 gives 60 words. Of these, 6 words don't contain an A in them anywhere (3*2*1). Of the remaining 54 words, 18 contain two As and 36 contain one A. Now remove the labels on the two As. The 6 words without any As are distinct. The remaining 54 words are double-counted. Pete, you only addressed the 18 words with 2 As. The 36 words with one A are also double-counted (A1BC and A2BC become one word upon removing the labels). 6+54/2 = 33, which is the same answer Zeno got.

Thanks DH :o

I think the answer is 33.

There are 9 cases which have 2 As.
AAB
ABA
BAA
AAC
ACA
CAA
AAD
ADA
DAA

There are 24 cases (4*3*2) for the rest of the possible words, some including only 1 A.
So, 24+9 = 33.

It's 60 words only if all of the letters are different.
Is there a way to find the number of arrangements with exactly 2 A's without having to LIST them all out?


Also, I am wondering if there is a way to find the number of arrangements with exactly 1 A.


Thanks!

Did you see my last post?

Did you see my last post?

Yes, thanks for your help, but it seems complicated to me...

How about this analysis?

There are 6 possible 3-letter words using only BCD.

For each of the above 6 words, you can make 3 more by substituting A for one of the other letters (The A could replace either the first, the second, or the third letter). This gives 18 more words.

For each of the above 6, you can substitue two A's in three ways (leave the first, second, or third letter as is and put A's in the other two places). This gives another 18 possibilities.

6 + 18 + 18 = 4260 = 5*4*3 is obviously silly. That is the number of words if you had 5 different letters. I am too lazy to analyse the other solutions.

For each of the above 6, you can substitue two A's in three ways (leave the first, second, or third letter as is and put A's in the other two places). This gives another 18 possibilities.
This gives you two of each two A word, so it actually only gives you another 9 possibilities, for a total of 33.

For example:
BCD gives AAD, ACA, BAA
But...
CBD also gives AAD,
DCB also gives ACA, and
BDC also gives BAA.

Pete: In another thread I said that it is easy to make mistakes when dealing with probability, permutations, combinations, & random processes.

I further said that I have a lot of experience and still mess up more often that I like to admit.

I am surprised that I messed this one up. It really is a simple analysis.

BTW: One reason why many people dislike and avoid mathematics & the hard sciences, is that it is so easy to be shown that you are wrong. In economics, sociology, psychology, political science (an oxymoron), et cetera you can stubbornly defend your opinions with obfuscating verbiage and nobody can clearly show you are wrong.

Combinations with identical elements:
2) You have 2 blue balls, 3 yellow balls, and 3 red balls in a box. If you randomly take 3 balls from the box, what is the probability that you have 1 ball of each color? (assume order does not matter)

This is even crazier, how can I even get a starting point for this problem? I know the definition of combination, but I still have absolutely no clue on solving this problem...

Any help is appreciated!

One way to solve a problem where order doesn't matter is to solve the related set of problems where order does matter. The probability of flipping heads/tails in order is 1/4, as is the probability of flipping tails/heads. Heads/tails and tails/heads are mutually exclusive events, so you can add the probabilities to yield the order-insensitive probability of flipping a heads and a tails. Some questions:
How many different ways are there to draw a blue ball, a yellow ball, and then a red ball, in any order?
This question is not related to the first question (yet): What is the probability of drawing a blue ball, a yellow ball, and then a red ball, in that specific order?
Now generalize: What is the probability of drawing a blue ball, a yellow ball, and then a red ball, in any other given specific order?
Now tie it all together: What is the probability of that the first three balls taken from the box are blue, yellow ball, and red?

One way to solve a problem where order doesn't matter is to solve the related set of problems where order does matter. The probability of flipping heads/tails in order is 1/4, as is the probability of flipping tails/heads. Heads/tails and tails/heads are mutually exclusive events, so you can add the probabilities to yield the order-insensitive probability of flipping a heads and a tails. Some questions:
How many different ways are there to draw a blue ball, a yellow ball, and then a red ball, in any order?
This question is not related to the first question (yet): What is the probability of drawing a blue ball, a yellow ball, and then a red ball, in that specific order?
Now generalize: What is the probability of drawing a blue ball, a yellow ball, and then a red ball, in any other given specific order?
Now tie it all together: What is the probability of that the first three balls taken from the box are blue, yellow ball, and red?


i) (2C1 x 3C1 x 3C1) x 6 ?
I multiplied by 6 since there are 6 mutually exclusive events, they are:
RBY (first ball red, second ball blue, third ball yellow)
RYB
BYR
BRY
YBR
YRB

ii) (2C1 x 3C1 x 3C1) / (8C3) ?
But wait a second, should they be permutations since it's in a "specific order"?


iii) (2C1 x 3C1 x 3C1) x 6 / (8C3) ?

iv) No clue...

Consider the possible ways to draw and the probability for each.BYR = (2/8)*(3/7)*(3/6)
BRY = (2/8)*(3/7)*(3/6)
YBR = (3/8)*(2/7)*(3/6)
YRB = (3/8)*(3/7)*(2/6)
RBY = (3/8)*(2/7)*(3/6)
RYB = (3/8)*(3/7)*(2/6)The above are mutually exclusive so we add them to get the total probability. Each of the above evaluates to 18/336

6*(18/336) = 0.321 429

A different method.There are 56 combinations of 8 objects taken 3 at a time (8*7*6 / 3*2*1)


How many combinations fit the conditions: BiYjR k ?


The above looks like 2*3*3 = 18


Probability = 18 / 56 or 0.321 429The above seems reasonable, but I would not bet a large sum of money on its being correct.

2) Is this correct?
2C1 x 3C1 x 3C1
----------------
8C3