specific combinations

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rize1159

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May 7, 2011
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I have fourteen characters like
a,b,c,d,e,f,g,h,i,j,k,l,m,n

now I want the combination/permutation in such a way that combination of same characters are not repeated i.e.
abcde=cdabe=ebacd etc, so only single combination is possible for same letters. Now question is that how many combination can be made for 14 letters and each combination consists of only 5 letters. and what would be the formula for this :D :lol:
 
rize1159 said:
I have fourteen characters like
a,b,c,d,e,f,g,h,i,j,k,l,m,n

now I want the combination/permutation in such a way that combination of same characters are not repeated i.e.
abcde=cdabe=ebacd etc, so only single combination is possible for same letters. Now question is that how many combination can be made for 14 letters and each combination consists of only 5 letters. and what would be the formula for this :D :lol:
Click to expand...
Your question is slightly ambiguous.

Is aaaaa an acceptable combination? I am going to assume that it is NOT.

Let's try an example. Let's start with selecting 2 characters out of 5.

How many ways are there to select 2 out of 5 characters if no character is to be repeated and different sequences of the same characters are to be treated as the same?

Well, there are 5 ways to select the first character. For each one of those 5 ways, there are 4 ways to select the second, right?

So, there are (5 * 4) = 20 ways to select without replacement a sequence of 2 items from 5 items. This is called the number of permutations.

Of course, this computation of the number of permutations considers (a, b) and (b, a) to be different. Consequently, we have overcounted if we are not concerned about sequence. How many ways can we sequence two items? Well, either a comes first and b comes second, or b comes first and a comes second. The number of possible sequences for each selection is 2. So, the correct answer in this case is (20 / 2) = 10. This is called the number of combinations. LET'S CHECK OUR ANSWER. In selecting (without replacement and disregarding order) 2 out of the 5 characters a, b, c, d, e, the possibilities are (a, b), (a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, d), (c, e), and (d, e). Sure enough, there are 10 combinations.

If you understand this one example, you will understand about permutations and combinations.

Now YOU calculate the number of permutations and combinations of selecting three items from a collection of five items.
First, what is the number of possible permutations?
Second, how many ways can you sequence each selection?
Third, what is the number of possible combinations?
 
rize1159 said:
I have fourteen characters like
a,b,c,d,e,f,g,h,i,j,k,l,m,n

now I want the combination/permutation in such a way that combination of same characters are not repeated i.e.
abcde=cdabe=ebacd etc, so only single combination is possible for same letters. Now question is that how many combination can be made for 14 letters and each combination consists of only 5 letters. and what would be the formula for this :D :lol:
Click to expand...

We designate the combinations of n things taken n at a time as nCn and the combinations of n things taken r at a time by nCr. To find the number of combinations of n dissimilar things taken r at a time, the formula is nCr = n!/[r!(n-r)!] which can be stated as n factorial divided by the product of r factorial times (n-r) factorial. Example: In how many ways can a committee of three people be selected from a group of 12 people? We have 12C3 = (12!)/[3!(9!) = 220. How many different ways can you combine A, B, C, and D in sets of three? Clearly, 4C3 = (4x3x2x1)/(3x2x1)(1) = 4. How many handshakes will take place between six people in a room when they each shakes hands with all the other people in the room one time? Here, 6C2 = (6x5x4x3x2x1)/(2x1)(4x3x2x1) =15.


Now, how many combination can be made of 14 letters where each combination consists of only 5 letters.

Your turn.
 
So as per your answers, answer would be 1001, 14 letters and combination of 4 not five.
Now what is the possibility of combination if 5th letter is a repeating one i.e aabcd, aabce, aabde, or abbcd, abbde, abbce etc.
pls answer in a question as a form of equation.
 
Well, we're not here to do your homework; read "Read before posting".
Show your work; use google for assistance...
 
Dear I am a thirty years oldman and trying to solve a puzzle so I am asking question step by step to make it easy to understand. If you want to know the whole puzzle at once, let me know.
 
rize1159 said:
Dear I am a thirty years oldman and trying to solve a puzzle so I am asking question step by step to make it easy to understand. If you want to know the whole puzzle at once, let me know.
Click to expand...

Rize There is a puzzle page (odds and ends) so the huge majority of questions asked on THIS page are homework problems. It is also helpful to the volunteers, as is explained under "Read Before Posting," to give some clue as to context and your degree of expertise.

I am not helping students at the moment so I can help you with your puzzle. The number of distinct sequences that you can form by making n selections without replacement from a set of n distinct things is n! = n * (n - 1) * (n - 2) * ... (2) * 1. You can pick the first n ways, the second (n - 1) ways, etc. By sequence, I mean a,b is different from b,a.

The number of distinct sequences that you can form by selecting k distinct things without replacement from a set of n distinct things is [n! / (n - k)!]. Here's why. You can pick the first in the sequence n ways, the second (n - 1) ways, etc. but you are taking only k items so only the first k terms in n! are relevant. You eliminate from n! the product associated with the (n - k) items not selected, which equals (n - k)!. In your first question, the number of possible sequences is [14! / (14 - 5)!] = (14! / 9!) = 14 * 13 * 12 * 11 * 10 = 240,240, but that is not your answer because you want to count different sequences that contain the same elements as 1 combination. Go back to the previous paragraph. The number of ways that you can sequence k things is k!. So the GENERAL answer to selecting k items from a set of n without replacement or regard to order is {n! / [(n - k)! * k!]}.

The answer to your original question is [240,240 / (5 * 4 * 3 * 2 * 1)] = (240,240 / 120) = 2,002. However, the problem seems to have shifted because suddenly we are talking about 4, not 5. 14! / [(14 - 4)! * 4!] = [(14 * 13 * 12 * 11) / (4 * 3 * 2 * 1) = (24,024 / 24) = 1,001. I have no idea whether that is right or not because I do not know why you jumped from selections of 5 to selections of 4.

I simply do not understand your final question. Is only 1 repeat allowed, or 1, 2, 3, 4, or 5 repeats? Are you still not concerned about sequences? That is, is aabbb to be considered the same as babab? The kinds of questions that you seem now to be asking do not have standard formulas. You need to work them out along the same lines as the formulas used for the common questions. The needed equations are very specific to the question being asked, and the question being asked must be stated very precisely. When repeats are allowed, you are selecting with replacement. The formulas given above assumed no replacement. The equations get MUCH messier when there is selection with replacement.
 
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