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

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Vipin Gupta

Last answered:

30 Nov 2021

Posted on:

13 Jun 2021

Need help for problem 5 given in exercise.

in Probability / Practical Example - Combinatrics

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Hristina Hristova

Posted on:

30 Nov 2021

Need help for problem 5 given in exercise.

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