Combinatorics and counting

2 free worked examples

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1. Exam PCombinatorics and countingCore

   An actuarial department must send 4 of its 11 analysts to a conference. In how many ways can the group be chosen?

   1. A$44$
   2. B$110$
   3. C$330$
   4. D$1,320$
   5. E$7,920$

   Solution

   1. The group is unordered - sending analysts A, B, C, D is the same as sending D, C, B, A - so this is a combination.
   2. $\left(\genfrac{}{}{0ex}{}{11}{4}\right)=\frac{11!}{4!7!}$.
   3. $=\frac{11\times 10\times 9\times 8}{4\times 3\times 2\times 1}=\frac{7920}{24}=330$.
   4. The ordered count $P(11,4)=7920$ is exactly $4!=24$ times too large, which is why it appears as a distractor.

   Trap. Using the permutation count 7,920, which treats two identical groups in different orders as different selections.

2. Exam PCombinatorics and countingCore

   A password consists of 3 letters (from 26, repeats allowed) followed by 2 digits (from 10, repeats allowed). How many passwords are possible?

   1. A$15,600$
   2. B$676,000$
   3. C$1,404,000$
   4. D$1,757,600$
   5. E$11,232,000$

   Solution

   1. Each position is chosen independently, so the multiplication rule applies directly.
   2. The letters contribute $26^3=17,576$.
   3. The digits contribute $10^2=100$.
   4. $17,576\times 100=1,757,600$ passwords.

   Trap. Using permutations without repetition when the problem explicitly allows repeats.