A class test consists of 4 Algebra questions,W ,X,Y and Z, and 3 Geometry questions,A,B,C and D.The teacher decides that the questions should be arranged in two sections.Algebra followed by Geometry,with questions in each section arranged in a random order Find the number of arrangement in which questions W and B are next to each other. Find the number of arrangements in which questions X and D are separated by more than four other subjects.
To find the number of arrangements in which questions W and B are next to each other, we can treat them as a single block. This means that instead of having 4 Algebra questions and 3 Geometry questions, we now have 3 "blocks" - the block with questions W and B as one, and the other two blocks remaining as Algebra and Geometry questions.
Let's find the number of arrangements for each block and multiply them together to get the final result:
1. The number of arrangements for the block with questions W and B:
Since W and B need to be next to each other, we can treat them as a single entity. This block has 2 questions, and we can arrange them in 2! = 2 factorial ways.
2. The number of arrangements for the Algebra block (excluding W and B):
We have 4 - 2 = 2 remaining Algebra questions (X and Y) in this block. We can arrange them in 2! ways.
3. The number of arrangements for the Geometry block:
We have 3 Geometry questions (A, C, and D) in this block. We can arrange them in 3! ways.
To find the total number of arrangements:
Total arrangements = (Number of arrangements for the block with W and B) * (Number of arrangements for the Algebra block) * (Number of arrangements for the Geometry block)
Total arrangements = 2! * 2! * 3!
Now let's calculate this:
Total arrangements = 2 * 2 * 3 * 2 * 1 = 24
So, there are 24 arrangements in which questions W and B are next to each other.
Now let's move on to the second part of the question.
To find the number of arrangements in which questions X and D are separated by more than four other subjects, we need to consider the possible number of subjects separating them.
If there are 5 or more subjects between X and D, we can arrange them in various ways.
1. Let's first consider 5 subjects between X and D:
There are 5 Algebra questions (W, X, Y, Z, and one more) and 3 Geometry questions (A, B, and C) that can be placed between X and D.
The number of ways to arrange these 8 questions is 8!.
However, X and D can swap positions, so we need to multiply by 2 to account for this possibility.
The number of arrangements with 5 subjects between X and D = 8! * 2.
2. Similarly, we can consider 6, 7, and 8 subjects between X and D:
The number of arrangements with 6 subjects between X and D = 9! * 2.
The number of arrangements with 7 subjects between X and D = 10! * 2.
The number of arrangements with 8 subjects between X and D = 11! * 2.
To find the total number of arrangements:
Total arrangements = (Number of arrangements with 5 subjects) + (Number of arrangements with 6 subjects) + (Number of arrangements with 7 subjects) + (Number of arrangements with 8 subjects)
Total arrangements = (8! * 2) + (9! * 2) + (10! * 2) + (11! * 2)
Now let's calculate this:
Total arrangements = 2 * (8! + 9! + 10! + 11!)
Using a calculator or software, you can find the numerical value of this expression.
So, the total number of arrangements in which questions X and D are separated by more than four other subjects depends on the calculations explained above.
To find the number of arrangements in which questions W and B are next to each other, we can treat W and B as a single entity. So, we have WB and XYZCDA as the remaining entities.
Now, we need to find the number of arrangements considering W and B as a single entity. This can be done by treating WB as a single entity along with XYZCDA. So, we have WBXYZCDA.
Now, the total number of arrangements for WBXYZCDA is 7!
However, within this arrangement, WB can be arranged in 2! ways.
So, the total number of arrangements in which W and B are next to each other is 7! x 2! = 10,080.
To find the number of arrangements in which questions X and D are separated by more than four other subjects, we need to consider two cases:
Case 1: When X is before D
In this case, we have the sequence X _ _ _ D _ _ _. We need to arrange W, Y, Z, A, B, C in the remaining blanks.
The total number of arrangements for W, Y, Z, A, B, C is 6!.
However, within this arrangement, X, D, and the remaining subjects can be arranged in 3! ways.
So, the total number of arrangements in this case is 6! x 3! = 2,160.
Case 2: When D is before X
In this case, we have the sequence D _ _ _ X _ _ _. Again, we need to arrange W, Y, Z, A, B, C in the remaining blanks.
Similar to Case 1, the total number of arrangements in this case is 6! x 3! = 2,160.
Therefore, the total number of arrangements in which questions X and D are separated by more than four other subjects is 2,160 + 2,160 = 4,320.