-
ithamarsorek
- Junior | Next Rank: 30 Posts
- Posts: 26
- Joined: Tue Dec 07, 2010 1:01 pm
Does anyone have a simpler explanation? I got most of it but trying to really understand expecially the part of dividing everything by 2 towards the end.
Thanks much and happy holiday!
In how many different ways can the letters A, A, B, B, B, C, D, E be arranged if the letter C must be to the right of the letter D?
1,680
2,160
2,520
3,240
3,360
Analyze the Question:
If you see a group and you are asked to arrange items in the group, you should immediately expect to calculate permutations or combinations.
Identify the Task:
Remember that with permutations order matters. If order does not matter, then we are dealing with combinations. We must use these formulas to find the number of possible orderings of the given letters, remembering to account for the "C is to the right of D" requirement.
Approach Strategically:
We have a total of 8 letters. Some of the letters appear more than once. The requirement of the question stem that the letter C is to be to the right of the letter D, with the letters D and C each appearing once.
If n is a positive integer, the number of different ways to arrange n different objects is n!, where n! is the product of the first n positive integers. Thus, n! = n (n − 1)(n − 2) ... (3)(2)(1).
If we had 8 different letters, the number different of ways to arrange the 8 letters would be 8! = (8)(7)(6)(5)(4)(3)(2)(1). However, this number will be lower because of duplicates in this problem. Because the letter A appears twice, we must divide 8! by 2! Because the letter B appears 3 times, we must divide 8! by 3!. So the number of different ways to arrange the letters A, A, B, B, B, C, D, E is . We also have the requirement that the letter C is to the right of the letter D. For any specified locations of the letter C and the letter D, there are the same number of arrangements for the other 6 letters when C is to the right of D and D is to the right of C. So we must divide the number of possible arrangements of all the letters by 2. So the number of different ways to arrange the letters A, A, B, B, B, C, D, E where the letter D is to the right of the letter C is , which is . To find the answer to the question, let's find the value of . We have = = = (4)(7)(3)(5)(4) = 1,680. Answer Choice (A) is correct.
Confirm your Answer:
Remember if you are canceling out factorials in a fraction, you need to clearly mark what is left
Thanks much and happy holiday!
In how many different ways can the letters A, A, B, B, B, C, D, E be arranged if the letter C must be to the right of the letter D?
1,680
2,160
2,520
3,240
3,360
Analyze the Question:
If you see a group and you are asked to arrange items in the group, you should immediately expect to calculate permutations or combinations.
Identify the Task:
Remember that with permutations order matters. If order does not matter, then we are dealing with combinations. We must use these formulas to find the number of possible orderings of the given letters, remembering to account for the "C is to the right of D" requirement.
Approach Strategically:
We have a total of 8 letters. Some of the letters appear more than once. The requirement of the question stem that the letter C is to be to the right of the letter D, with the letters D and C each appearing once.
If n is a positive integer, the number of different ways to arrange n different objects is n!, where n! is the product of the first n positive integers. Thus, n! = n (n − 1)(n − 2) ... (3)(2)(1).
If we had 8 different letters, the number different of ways to arrange the 8 letters would be 8! = (8)(7)(6)(5)(4)(3)(2)(1). However, this number will be lower because of duplicates in this problem. Because the letter A appears twice, we must divide 8! by 2! Because the letter B appears 3 times, we must divide 8! by 3!. So the number of different ways to arrange the letters A, A, B, B, B, C, D, E is . We also have the requirement that the letter C is to the right of the letter D. For any specified locations of the letter C and the letter D, there are the same number of arrangements for the other 6 letters when C is to the right of D and D is to the right of C. So we must divide the number of possible arrangements of all the letters by 2. So the number of different ways to arrange the letters A, A, B, B, B, C, D, E where the letter D is to the right of the letter C is , which is . To find the answer to the question, let's find the value of . We have = = = (4)(7)(3)(5)(4) = 1,680. Answer Choice (A) is correct.
Confirm your Answer:
Remember if you are canceling out factorials in a fraction, you need to clearly mark what is left
