Question: Alice attends a small college in which each class meets only once a week. She is deciding between 30 non-overlapping classes. There are 6 classes to choose from for each day of the week, Monday through Friday. Trusting in the benevolence of randomness, Alice decides to register for 7 randomly selected classes out of the 30, with all choices equally likely. What is the probability that she will have classes every day, Monday through Friday? (This problem can be done either directly using the naive definition of probability, or using inclusion-exclusion.)

Answer as per me: Naive probability approach: Total choices are $^{30}{C}_7$. There are buckets with 6 classes each for each of the five days. So we will need to choose one from these buckets. Then from the 25 choices left choose random 2. Prob: $\frac{(6^5.^{25}{C}_2)}{({30}{C}_7)}$ This is greater than 1, so there is something obviously wrong

Actual answer: There are two general ways that Alice can have class every day: either she has 2 days with 2 classes and 3 days with 1 class, or she has 1 day with 3 classes, and has 1 class on each of the other 4 days. Using this they calculate.

Please guide on what was wrong with my approach

• Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Aug 21 '18 at 15:13
• Thanks have updated appropriately. Hope this is fine – gooner Aug 21 '18 at 15:20
• Yes, I think so. – José Carlos Santos Aug 21 '18 at 15:28
• An error in your approach is that you have counted some sets of 7 choices at least twice. If I make 5 choices followed by 2 more choices, and you do the same, my set of 5 initial choices may differ from yours, but we may finish with the same set of 7 choices. Your approach would count these as different sets of 7. – DanielWainfleet Aug 21 '18 at 15:36
• Consider a much smaller problem where you have only one bucket and it has two balls in it, you take two balls. You ask what the probability that every bucket (all one of them) were taken from. According to your method, "pick which appears in the first bucket" $2$ choices, then "pick what else appears in the bucket", $1$ choices. and this is out of the $\binom{2}{2}=1$ ways of picking things, giving a supposed probability of $\dfrac{2\cdot 1}{1}$. Notice, in the numerator it supposedly mattered what order they were picked when in reality it shouldn't have mattered. – JMoravitz Aug 21 '18 at 15:36

The problem with your approach is the typical one;: you are counting the same thing more than once. Let $k\in\{1,2,3,4,5\}$ and suppose that the contents of the $k$th bucket consists of $\{a_k,b_k,c_k,d_k,e_k\}$. Then Alice could choose the $a$'s from all buckets, together with $b_1$ and $c_1$. But this single possibility is counted as $3$ different possibilities using your method of counting, since you could group $b_1$ and $c_1$ together, you could group $a_1$ and $c_1$ together, and you could group $a_1$ and $b_1$ together.
• Oh, essentially, $a_1, b_1, c_1, d_1, e_1$ and $a_2$ from from the $^{25}C_2$ were considered different from $a_2, b_1, c_1, d_1, e_1$ and $a_1$ from the $^{25}C_2$ in my approach. Hence the double counting right. I got the point. Much appreciated – gooner Aug 21 '18 at 15:41