This question has already been answered but I was curious as to why the answer I came up with is wrong. What is wrong with my logic here.

$$\frac{7! \binom{12}{7} 7^5}{7^{12}}$$

Where $\binom{12}{7}$ is the number of ways to choose the 7 calls that will be assigned to each day (guaranteeing each day has a minimum of one call) and $7!$ represents the number of ways to order those calls and $7^5$ represents the number of ways to distribute the five remaining calls. What am I getting wrong here?


You are counting each distribution of calls multiple times. Say the calls are distributed as follows:

\begin{array}{c c c c c c c} \text{Sunday} & \text{Monday} & \text{Tuesday} & \text{Wednesday} & \text{Thursday} & \text{Friday} & \text{Saturday}\\ \hline c_1 & c_2, c_3 & c_4 & c_5, c_6, c_7 & c_8 & c_9, c_{10} & c_{11}, c_{12} \end{array} You count this scenario $1 \cdot 2 \cdot 1 \cdot 3 \cdot 1 \cdot 2 \cdot 2 = 24$ times, once for each way you could designate one of the calls you receive each day as the call you receive that day and the remaining calls as being among the additional five calls you receive that week.


You already have an explanation why the way to count does not work.

Here is one way you can count. First distribute 7 calls one on each day, then however you place the rest the condition will be fulfilled. You have 5 calls left to place. You can place them any way among 7 days. So you got two components right, but in wrong places:


If I got it right it will be $\approx$ 4.7123 %. Simulating $10^7$ weeks in Octave give me 4.7031% of the time.

  • 1
    $\begingroup$ Just saw this because I'm doing the same problem, but this answer is actually incorrect. The correct probability is $\approx$ 22.85 %. $\endgroup$ – Ryker Feb 8 at 21:35
  • $\begingroup$ Finally someone slightly awake. $\endgroup$ – mathreadler Feb 8 at 21:44

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