How many possible arrangements Permutations

In summary, the person is new to the forum and seeking help with two problems related to arranging letters. The first problem involves arranging six letters, with the condition that b and c must be next to each other. The second problem involves the same six letters, but with the condition that b and c cannot be next to each other. The speaker suggests starting with an easier version of the problem and noticing a pattern.
  • #1
Camel
1
0
Hi guys, I'm new in this forum, so i hope you can help me with these problems. It might be easy, but i just started taking Elementary stats... so here they are...

a) how many possible arrangements are there of the six letters {a,b,c,d,e,f} in which b and c are next to each other?

b) how many possible arrangements are there of the six letters {a,b,c,d,e,f} in which b and c are not next to each other?

thanx for your help in advance
 
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  • #2
Perhaps you could start with an easier version of the problem -
How many ways are there for b,c?
What about a,b,c? a,b,c,d?
You might notice a pattern.
 
  • #3


Hello and welcome to the forum! I'd be happy to help with your questions about permutations.

a) To find the number of possible arrangements where b and c are next to each other, we can treat them as one "block" of letters. This means we have 5 letters to arrange: {a, (bc), d, e, f}. The number of arrangements of 5 letters is 5!, but since b and c can also be arranged within their block, we need to multiply by 2!. This gives us a total of 5! * 2! = 240 possible arrangements.

b) To find the number of possible arrangements where b and c are not next to each other, we can first find the total number of arrangements of all 6 letters, which is 6!. Then, we can subtract the number of arrangements where b and c are next to each other that we found in part a) (240). This gives us a total of 6! - 240 = 720 - 240 = 480 possible arrangements.

I hope this helps! Let me know if you have any further questions. Good luck with your stats studies!
 

1. How many possible arrangements can be made with a set of n objects?

The number of possible arrangements, or permutations, of a set of n objects is n!, where n! represents the factorial of n. This means that for every object in the set, there are n options to choose from for the first position, n-1 options for the second position, and so on until all n positions are filled.

2. How does the number of possible arrangements change if some objects are repeated?

If some objects in the set are repeated, the number of possible arrangements decreases. For example, if there are n objects in the set and two of them are identical, the number of possible arrangements is reduced by a factor of 2. This is because the identical objects can be switched without changing the overall arrangement.

3. Can the order of the objects in a permutation affect the total number of arrangements?

Yes, the order of the objects in a permutation can greatly affect the total number of arrangements. For example, the number of possible arrangements of the letters in the word "apple" is 5!, but if the letters are rearranged to spell "pepla", the number of possible arrangements is reduced to 5!/2, since the two "p" letters are identical.

4. How do permutations differ from combinations?

Permutations and combinations are both ways of counting the number of possible arrangements, but they differ in how they treat the order of the objects. Permutations take into account the order of the objects, while combinations do not. This means that in a combination, "ABC" and "CBA" would be considered the same arrangement, while in a permutation, they would be considered different arrangements.

5. In what real-life situations can permutations be used?

Permutations have many practical applications in fields such as mathematics, computer science, and statistics. They can be used to calculate probabilities, analyze data, and solve problems related to ordering and arranging objects. For example, in a game of cards, the number of possible arrangements of a deck can be calculated using permutations.

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