1. The problem statement, all variables and given/known data It is decided to label the vertices of a rectangle with the letters A, B, C and D. In how many ways is this possible if: a) they are to be in clockwise alphabetical order b) they are to be in alphabetical order c) they are to be in random order 2. Relevant equations 3. The attempt at a solution How do I solve for c? What is the logic behind C?
You've done (a) and (b)? For (c), label the vertices 1, 2, 3, 4. How many different choices of A, B, C, and D are there for vertex 1? After you have done that, how many choices are there for vertex 2? Once 1 and 2 are labled, how many choices are there for vertex 3? Finally, how many for vertex 4? Now use the "Product Principle".
I`m now struggling with the same problem . Can you pls show working for part a and b as well ? As the principle product stated , so the answer for a must be 8 but in the answer key it's 4 ??? Can you explain why ?
Then you are using the principle wrong. The "product principle" (also called the "fundamental principle of counting") says if there are n ways of doing one thing and b ways of doing another there are nm ways of doing both. In (a) you are required to label the vertices clockwise. Okay, how many choices do you have for vertex "A"? Now, you have NO MORE CHOICES! After you have chosen where to put the "A", the vertex clockwise from "A" MUST be labeled "B" and so on. The difference between (a) and (b) is that in (b), after you have chosen which vertex to label "A", you still have to choose whether to go clockwise or counterclockwise.
I can see it quite easily in my mind but like the theory to get it seems quite complex . According to the concept of the product of principle . So there are 4 ways to label A , either is B,C,R . Then isn't it supposed to be 4 x 4 x 4 x 4 ?
^^ I actually just got it myself . If we think geometrically like for clockwise u can have 4 choices for labeling the first vertex then for the next 3 vertexes u can only have one choice for each . Therefore it's 4 for a , 8 for b and for c . I have 4 choices for vertex A , 3 for vertex B and 2 For C , 1 For D ==> 24 . Correct compared with the answer key . Can you help me with the next question How many 3-digit numbers can be formed using the digits 2,3,4,5 and 6 (a) as often as desired . (b) once only ? And can you explain why I have 42 for this question :In how many different ways can the top two positions be filled in a table tennis competition of 7 teams ? I got 42 because i take 7 x 6 ( i got this way of solution when learn about permutation but still haven't got why is it 7 x 6 ? )