aA 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 W and B as a single entity.
1. Treat W and B as a single entity, call it WB.
2. Now we have 6 entities: X, Y, Z, WB, A, C, D.
3. Arrange these 6 entities in the two sections (Algebra and Geometry), where questions in each section are arranged in a random order.
There are 2 sections, so we have 2 choices for the first section.
Once we select the first section, we have 6 entities to arrange in that section.
Therefore, the number of arrangements for the first section is 6! (factorial of 6).
The remaining 4 entities will go to the second section.
Therefore, the number of arrangements for the second section is 4! (factorial of 4).
So the number of total arrangements is 2 * 6! * 4!.
To find the number of arrangements in which questions X and D are separated by more than four other subjects, we can consider the following cases:
Case 1: X is before D
1. There are 4 subjects that can be placed before X (W, Y, Z, WB).
2. There are 4 subjects that can be placed between X and D (A, C, and the two remaining subjects from Algebra section).
Therefore, the number of arrangements for Case 1 is 4 * 4! * 3! * 3! (4 choices for placement before X, factorial of 4 for arrangement of remaining Algebra subjects, factorial of 3 for arrangement of subjects between X and D, and factorial of 3 for arrangement of Geometry subjects).
Case 2: D is before X
1. There are 6 subjects that can be placed before D (W, X, Y, Z, WB, and one subject from Geometry section).
2. There are 2 subjects that can be placed between D and X (A and the remaining subject from Geometry section).
3. There are 2 subjects that can be placed after X (one from Algebra section and C).
Therefore, the number of arrangements for Case 2 is 6 * 2 * 2! * 2! (6 choices for placement before D, 2 choices for placement between D and X, factorial of 2 for arrangement of remaining Algebra subjects, and factorial of 2 for arrangement of subjects after X).
To get the total number of arrangements satisfying the condition, we add the number of arrangements from Case 1 and Case 2.
Total number of arrangements = (4 * 4! * 3! * 3!) + (6 * 2 * 2! * 2!)
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 WWB as one unit. Now, we have 3 units: WWB, X, YZ, and ACD.
The number of arrangements of these units is 3!, which is equal to 3 factorial. This is because we have 3 units, and the order in which we arrange them matters. The factorial of a number represents the product of all positive integers less than or equal to that number.
Therefore, the number of arrangements in which questions W and B are next to each other is 3!.
To find the number of arrangements in which questions X and D are separated by more than four other subjects, we can consider X and D as two separate units. We can place these two units in any of the available positions between the other units.
Since there are two sections (Algebra and Geometry), we have two potential locations where we can place X and D: between WWB and ACD or between ACD and YZ.
In the first case, we have 4 possible positions to place X and D: X D, XX D, X DX, or XX DX. Similarly, in the second case, there are 4 possible positions as well.
Hence, the total number of arrangements in which questions X and D are separated by more than four other subjects is 4 + 4 = 8.