In problem 5.a, why can't we have 5*5(if repetition allowed or else 5*4) as probable number of cake combinations ? First person can have any choice out of 5 and so for the second person. Why do I need to go for combination instead of calculating each position independently ?

Hey Vipin, Thank you for your question and apologies for the delayed answer! Let's denote all five cakes by the letters A through E. Note that our assumption in a) is that both twins enjoy all 5 cakes. Therefore, it doesn't matter whether: 1. Amy gets cake A and Steve gets cake B, or 2. Amy gets cake B and Steve gets cake A. The same would go if we pick cakes A and C, cakes A and D, cakes A and E and so on.. In total, there are 10 ways in which we can pick 2 cakes out of 5 if it doesn't matter which twin gets which cake. Now, imagine we are allowed to pick the same cake twice. Therefore, we have five additional options: picking two cakes of type A, two cakes of type B, ... , two cakes of type E. Therefore, combination with repetition would give us 10 + 5 = 15 ways in which we can pick two cakes if we are allowed to pick two of the same. Note that your way is completely valid once we put the personalized birthday sign because order is taken into consideration. Your way of solving the task distinguishes between these two cases: 1. Amy getting cake A and Steve getting cake B, or 2. Amy getting cake B and Steve getting cake A. I hope this helps! Kind regards, 365 Hristina

Wow, I was just about to ask the same question. from your answer; treating them as independent combinations makes it a variation. and given that. C=V/P with V=5*4 and P=2 therefore 20/2 = 10 anyways this is just my reasoning. my issue with this question 5 is that it's open to any kind of interpretation. "find the number of ways the cake can be ordered" could easily mean variation or combination.